Shortcuts & Speed Methods

Values from $a + b$ and $ab$ in One Line

Most two-letter identity questions give you two facts and ask for a third. Each method below gives the signal to look for, the fast way, one worked question and the mistakes it invites. In algebra those mistakes are small but costly: a minus sign dropped, the wrong square root taken, or the plain product used where the scaled one was needed.

When to use it: two of $a + b$, $a - b$, $ab$ and $a^2 + b^2$ are given, and some other value is asked.

Fast way: turn the data into $s = a + b$ and $p = ab$ first. After that, every answer is a formula you already know.

Asked	In terms of $s$ and $p$
$a^2 + b^2$	$s^2 - 2p$
$(a - b)^2$	$s^2 - 4p$
$a^3 + b^3$	$s^3 - 3ps$
$a^4 + b^4$	$(s^2 - 2p)^2 - 2p^2$
$\frac{a}{b} + \frac{b}{a}$	$\frac{s^2 - 2p}{p}$
$\frac{1}{a} + \frac{1}{b}$	$\frac{s}{p}$

If $d = a - b$ is given instead of the sum, the minus versions are $a^2 + b^2 = d^2 + 2p$ and $a^3 - b^3 = d^3 + 3pd$.

Question 1

If $x + y = 9$ and $xy = 14$, find $x^3 + y^3$ and $x^4 + y^4$.

Answer. Here $s = 9$ and $p = 14$. So $x^3 + y^3 = 729 - 3 \times 14 \times 9 = 729 - 378$ = 351. For the fourth powers, first $x^2 + y^2 = 81 - 28 = 53$, then $x^4 + y^4 = 53^2 - 2 \times 14^2 = 2809 - 392$ = 2417.

Time check: about 20 seconds.

Scaled letters: rename and reuse

When the data is $2x + 7y = 16$ and $xy = 4$, and the question asks for $8x^3 + 343y^3$, call $A = 2x$ and $B = 7y$. Then $A + B = 16$ and $AB = 14xy = 56$, so

$$8x^3 + 343y^3 = A^3 + B^3 = 16^3 - 3 \times 56 \times 16 = 4096 - 2688 = 1408$$

Going backwards

Sometimes the cubes are given and the sum is asked: $a^3 + b^3 = 341$ and $ab = 30$. Then $s^3 - 90s = 341$. Don't solve the cubic; try the options. $s = 11$ gives $1331 - 990 = 341$, so $a + b = 11$ (the numbers are 5 and 6).

When it does not work:

• Taking a square root. $(a - b)^2 = s^2 - 4p$ allows two signs. Use the one that fits the question; for $a > b$, take the positive root.
• Forgetting to scale the product. With $A = 2x$ and $B = 7y$, the product is $14xy$, not $xy$.
• The last term of $a^4 + b^4$ is $2p^2$, not $2p$.
• If $s^2 - 4p$ comes out negative, no real $a$ and $b$ fit the data. SSC questions are set so that this does not happen, so a negative value usually means a slip in your working.

Spot the Hidden Identity in an Algebraic Fraction

When to use it: a fraction with a long cubic or fourth-power expression on top, such as $\frac{8x^3 - 36x^2 + 54x - 27}{4x^2 - 9}$, and options that look much shorter.

Fast way: before you factorise anything, check the top against these patterns.

• Perfect cube. $x^3 - 3kx^2 + 3k^2x - k^3 = (x - k)^3$. The test: the last number is a cube $k^3$, and the two middle numbers are $3k$ and $3k^2$. With a number in front, as in $8x^3$, think of $(2x - k)^3$.
• Sum or difference of cubes over the matching bracket. $\frac{a^3 - b^3}{a^2 + ab + b^2} = a - b$ and $\frac{a^3 + b^3}{a^2 - ab + b^2} = a + b$.
• Fourth powers. $a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)$ and $a^4 - b^4 = (a^2 + b^2)(a + b)(a - b)$.
• Difference of squares. $\frac{9p^2 - 25q^2}{3p + 5q} = 3p - 5q$.

Question 2

Simplify $\frac{8x^3 - 36x^2 + 54x - 27}{4x^2 - 9}$.

Answer. The top is $(2x)^3 - 3(2x)^2(3) + 3(2x)(3^2) - 3^3$, which is $(2x - 3)^3$. Check the middle numbers: $3 \times 4 \times 3 = 36$ and $3 \times 2 \times 9 = 54$. The bottom is $(2x - 3)(2x + 3)$. Cancel one $(2x - 3)$:

$$\frac{(2x - 3)^3}{(2x - 3)(2x + 3)} = \frac{(2x - 3)^2}{2x + 3}$$

So the answer is $\frac{(2x - 3)^2}{2x + 3}$.

Time check: about 25 seconds.

When it does not work:

• The signs must follow the pattern. $(x - k)^3$ goes plus, minus, plus, minus, and $(x + k)^3$ is all plus. If one sign or one number is off, it is not a cube; factorise the usual way.
• $\frac{a^3 + b^3}{a^2 + ab + b^2}$ does not cancel. The middle sign below must be the opposite of the sign between the cubes.
• Don't cancel across a plus sign. In $\frac{x^2 + 9}{x + 3}$ nothing cancels, because $x^2 + 9$ has no real factors.

Put Simple Values In to Test the Options

When to use it: the options are expressions, not numbers, as in 'the expression simplifies to', 'one of the factors is' or 'which of these is equal to'. It also works with a condition such as $a + b + c = 0$, as long as your numbers obey it.

Fast way: pick small numbers that make the working trivial, such as $a = 1$ and $b = 0$, or $x = 1$. Work out the question with them, then each option. Any option that gives a different number is out. If two options survive, try a second pair, such as $a = 2$ and $b = 1$.

Question 3

$a^3 - b^3 - (a - b)^3$ is equal to (a) $3ab(a - b)$ (b) $3ab(a + b)$ (c) $-3ab(a - b)$ (d) $3a^2b$.

Answer. Put $a = 1$, $b = 0$: the question gives $1 - 0 - 1 = 0$, and every option gives 0 too, so this pair tells us nothing. Try $a = 2$, $b = 1$: the question gives $8 - 1 - 1 = 6$, and the options give 6, 18, $-6$ and 12. Only (a) survives.

Time check: about 30 seconds.

With a condition

If $a + b + c = 0$, take $a = 1$, $b = 1$, $c = -2$. Then $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = -\frac{1}{2} - \frac{1}{2} + 4 = 3$, and that is the answer for every such triple. If the condition is $a^2 + b^2 + c^2 = ab + bc + ca$, which forces $a = b = c$, put all three equal to 1.

When it does not work:

• The question asks for a number that depends on the data. Simple values only work for something true for all values (an identity), or for values that obey the given condition.
• Your values break the condition. For $x + \frac{1}{x} = 3$ you cannot put $x = 1$, because then $x + \frac{1}{x} = 2$. Use the table that comes next instead.
• Your values make a bottom zero. In $\frac{a + b}{a - b}$, never put $a = b$.
• 0 and 1 make many terms vanish, so several options can tie. Settle a tie with a second pair before you mark.

With the condition $a + b + c = 0$, the set 1, 1 and $-2$ is the quickest. Avoid 0 there, because questions with $abc$ in a bottom break down.

The $x \pm \frac{1}{x}$ Table

When to use it: $x + \frac{1}{x}$ or $x - \frac{1}{x}$ is given, and a higher power is asked.

Fast way: call $k = x + \frac{1}{x}$ and $m = x - \frac{1}{x}$. Then

$$x^2 + \frac{1}{x^2} = k^2 - 2 = m^2 + 2$$

$$x^3 + \frac{1}{x^3} = k^3 - 3k \qquad x^3 - \frac{1}{x^3} = m^3 + 3m$$

Higher powers come from these:

• Fourth power: square the second power and take 2 away, $x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2$, whichever family you start from.
• Fifth power: $x^5 + \frac{1}{x^5} = \left(x^2 + \frac{1}{x^2}\right)\left(x^3 + \frac{1}{x^3}\right) - \left(x + \frac{1}{x}\right)$.
• Sixth power: $x^6 + \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)^2 - 2$.
• Switching families: $k^2 - m^2 = 4$, so $x - \frac{1}{x} = \pm\sqrt{k^2 - 4}$, and $x^2 - \frac{1}{x^2} = km$.

Values worth knowing by heart:

$x + \frac{1}{x}$	$x^2 + \frac{1}{x^2}$	$x^3 + \frac{1}{x^3}$	$x^4 + \frac{1}{x^4}$
3	7	18	47
4	14	52	194
5	23	110	527
6	34	198	1154

Question 4

If $x - \frac{1}{x} = 5$, find $x^3 - \frac{1}{x^3}$.

Answer. It is the minus family, so use $m^3 + 3m = 125 + 15$ = 140. The trap option is 110, which is what the plus-family formula $k^3 - 3k$ gives.

Time check: about 10 seconds.

When it does not work:

• Mixing the families. The minus family has $+2$ and $+3m$; the plus family has $-2$ and $-3k$.
• $x^2 - \frac{1}{x^2}$ needs both $k$ and $m$, and the sign of $m$ depends on $x$. For $x > 1$, $x - \frac{1}{x}$ is positive; for $0 < x < 1$ it is negative.
• Going from $k$ to $m$ needs $k^2 \ge 4$. If $x + \frac{1}{x} = 1$, there is no real $x$; values like that are covered just below.

[SSC Tip] Going backwards is just as quick. If $x^4 + \frac{1}{x^4} = 47$, then $x^2 + \frac{1}{x^2} = \sqrt{49} = 7$, and for $x > 0$, $x + \frac{1}{x} = \sqrt{7 + 2} = 3$.

Divide by $x$: Turning Equations into $x + \frac{1}{x}$

When to use it: the question gives $x^2 - kx + 1 = 0$, $x^2 + 1 = kx$, $\frac{x}{x^2 + 1} = c$ or $\frac{x}{x^2 - kx + 1} = c$, or asks for an expression like $\frac{x^4 + 1}{x^2}$ or $\frac{x^6 + 1}{x^3}$.

Fast way: whenever the first and last terms are $x^2$ and 1, divide everything by $x$.

• $x^2 - 8x + 1 = 0$ becomes $x - 8 + \frac{1}{x} = 0$, so $x + \frac{1}{x} = 8$.
• $\frac{x}{x^2 + 1} = \frac{1}{6}$: turn both sides over, $\frac{x^2 + 1}{x} = 6$, so $x + \frac{1}{x} = 6$.
• $\frac{x^4 + 1}{x^2} = x^2 + \frac{1}{x^2}$ and $\frac{x^6 + 1}{x^3} = x^3 + \frac{1}{x^3}$.
• $x^2 - 7x - 1 = 0$ divides to $x - \frac{1}{x} = 7$, the minus family.

Forms like $ax + \frac{b}{x}$

Make the two terms reciprocals of each other first.

• $3t + \frac{1}{3t}$ is $y + \frac{1}{y}$ with $y = 3t$.
• $x + \frac{1}{4x} = 2$: multiply by 2 to get $2x + \frac{1}{2x} = 4$, so $4x^2 + \frac{1}{4x^2} = 16 - 2 = 14$.
• When the terms cannot be made reciprocals, square and watch the middle term: $\left(3x + \frac{1}{2x}\right)^2 = 9x^2 + \frac{1}{4x^2} + 3$.

Question 5

If $\frac{x}{x^2 - 3x + 1} = \frac{1}{2}$, find $x^3 + \frac{1}{x^3}$.

Answer. Turn it over: $\frac{x^2 - 3x + 1}{x} = 2$, so $x - 3 + \frac{1}{x} = 2$ and $x + \frac{1}{x} = 5$. Then $x^3 + \frac{1}{x^3} = 125 - 15$ = 110.

Time check: about 15 seconds.

When it does not work:

• The end terms must match. $x^2 - 5x + 4 = 0$ divides to $x + \frac{4}{x} = 5$, which is not the standard form; solve that one directly ($x = 1$ or $x = 4$).
• Turning over needs a single fraction on each side. Something like $\frac{x}{x^2 + 1} + 2 = 3$ must be tidied first.
• Keep the sign of the middle term. $x^2 + 5x + 1 = 0$ gives $x + \frac{1}{x} = -5$, and then $x^3 + \frac{1}{x^3} = -125 + 15 = -110$.

Special Values of $x + \frac{1}{x}$ and Cyclic Powers

When to use it: $x + \frac{1}{x}$ is one of 2, $-2$, 1, $-1$, 0, $\sqrt{2}$ or $\sqrt{3}$, or the equation is $x^2 + 1 = x$, $x^2 + x + 1 = 0$ or similar, and a huge power such as $x^{100}$ is asked.

Fast way: learn this table, then cut the power down.

Given	What it means
$x + \frac{1}{x} = 2$	$x = 1$
$x + \frac{1}{x} = -2$	$x = -1$
$x + \frac{1}{x} = 1$, that is $x^2 - x + 1 = 0$	$x^3 = -1$, so $x^6 = 1$
$x + \frac{1}{x} = -1$, that is $x^2 + x + 1 = 0$	$x^3 = 1$
$x + \frac{1}{x} = 0$	$x^2 = -1$, so $x^4 = 1$
$x + \frac{1}{x} = \sqrt{2}$	$x^4 = -1$, so $x^8 = 1$
$x + \frac{1}{x} = \sqrt{3}$	$x^6 = -1$, so $x^{12} = 1$

Where these come from: $x^2 - x + 1 = 0$ times $(x + 1)$ gives $x^3 + 1 = 0$. For $\sqrt{3}$, square it: $x^2 + \frac{1}{x^2} = 1$, so $x^4 - x^2 + 1 = 0$, and times $(x^2 + 1)$ this is $x^6 + 1 = 0$.

To cut a power down, use the cycle. If $x^3 = 1$, divide the power by 3 and keep only the remainder: $x^{100} = x^{99} \times x = x$.

Question 6

If $x + \frac{1}{x} = -1$, find $x^{100} + x^{50} + 1$.

Answer. Here $x^3 = 1$. Then $x^{100} = x^{99} \times x = x$ and $x^{50} = x^{48} \times x^2 = x^2$. So the expression is $x^2 + x + 1$, which is 0, because $x + \frac{1}{x} = -1$ means exactly $x^2 + x + 1 = 0$. The answer is 0.

Time check: about 20 seconds.

When it does not work:

• The sign flips. $x + \frac{1}{x} = 1$ gives $x^3 = -1$, not 1, and $x + \frac{1}{x} = -1$ gives $x^3 = 1$.
• $x^3 = -1$ here does not mean $x = -1$; that would make $x + \frac{1}{x} = -2$. These values of $x$ are not real numbers, but the algebra still holds, and SSC questions use it freely.
• For any other value of $x + \frac{1}{x}$ there is no short cycle. Use the table above.

With $x^3 = -1$, even multiples of 3 give $+1$ and odd multiples give $-1$: $x^{30} = 1$ but $x^{33} = -1$.

Conjugate Surds: $x + \frac{1}{x}$ Straight from the Surd

When to use it: $x$ is a surd like $17 + 12\sqrt{2}$ or $\sqrt{8} + \sqrt{7}$, and some power of $x$ and $\frac{1}{x}$ is asked.

Fast way: for $x = a + \sqrt{b}$, work out $a^2 - b$. If it is 1, then $\frac{1}{x} = a - \sqrt{b}$, so

$$x + \frac{1}{x} = 2a \qquad x - \frac{1}{x} = 2\sqrt{b}$$

and the rest comes from the table. Pairs worth knowing on sight: $2 \pm \sqrt{3}$, $3 \pm 2\sqrt{2}$, $5 \pm 2\sqrt{6}$, $7 \pm 4\sqrt{3}$, $9 \pm 4\sqrt{5}$ and $8 \pm 3\sqrt{7}$. For each, (first number)$^2$ minus (root part)$^2$ is 1; for $7 + 4\sqrt{3}$ it is $49 - 48$, so $x + \frac{1}{x} = 14$.

For $x = \sqrt{a} + \sqrt{b}$ with $a - b = 1$, the same idea gives $x + \frac{1}{x} = 2\sqrt{a}$. So $x = \sqrt{8} + \sqrt{7}$ gives $x + \frac{1}{x} = 2\sqrt{8} = 4\sqrt{2}$, and $x^2 + \frac{1}{x^2} = 32 - 2 = 30$.

For square roots of $x$, use

$$\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = x + \frac{1}{x} + 2 \qquad \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 = x + \frac{1}{x} - 2$$

Question 7

If $x = 17 + 12\sqrt{2}$, find $x^2 + \frac{1}{x^2}$ and $\sqrt{x} + \frac{1}{\sqrt{x}}$.

Answer. The root part squared is $(12\sqrt{2})^2 = 288$, and $289 - 288 = 1$, so $\frac{1}{x} = 17 - 12\sqrt{2}$ and $x + \frac{1}{x} = 34$. Then $x^2 + \frac{1}{x^2} = 1156 - 2$ = 1154. For the root, $\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = 34 + 2 = 36$, so $\sqrt{x} + \frac{1}{\sqrt{x}}$ = 6. This fits, because $\sqrt{x} = 3 + 2\sqrt{2}$: indeed $(3 + 2\sqrt{2})^2 = 17 + 12\sqrt{2}$.

Time check: about 20 seconds.

When it does not work:

• $a^2 - b$ is not 1. For $x = 3 + \sqrt{5}$, $9 - 5 = 4$, so $\frac{1}{x} = \frac{3 - \sqrt{5}}{4}$, and $x + \frac{1}{x}$ is not 6. If the question pairs $x$ with $y = 3 - \sqrt{5}$, use $x + y = 6$ and $xy = 4$, as in the first method of this section.
• The number in front of the root gets squared. For $5 + 2\sqrt{6}$, the root part squared is $(2\sqrt{6})^2 = 24$, not 6.
• When $x$ is less than 1, such as $9 - 4\sqrt{5}$, $x - \frac{1}{x}$ is negative: here it is $-8\sqrt{5}$.

For the pair $\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}$ and its upside-down form, see Chapter 2.

Use the Given Equation to Cut Down Powers

When to use it: $x$ is a surd such as $\sqrt{7} + 3$, or an equation such as $a^3 + 6a^2 + 12a + 3 = 0$ is given, and a long expression in $x$ or $a$ is asked.

Fast way: don't put the surd into the expression. First make the equation that $x$ satisfies, then use it to knock out the high powers.

• Get the root alone and square. $x = \sqrt{7} + 3$ gives $x - 3 = \sqrt{7}$, so $x^2 - 6x + 9 = 7$, that is $x^2 - 6x + 2 = 0$.
• Spot a cube. $a^3 + 6a^2 + 12a + 8 = (a + 2)^3$, so $a^3 + 6a^2 + 12a + 3 = 0$ means $(a + 2)^3 = 5$.
• For the square of a surd, just square it. $x = \sqrt{6} + 2$ gives $x^2 = 10 + 4\sqrt{6}$, so $x^2 - 4\sqrt{6} = 10$ straight away.

Question 8

If $x = \sqrt{7} + 3$, find $x^3 - 6x^2 + 2x + 9$.

Answer. From above, $x^2 - 6x + 2 = 0$. The first three terms are $x(x^2 - 6x + 2) = x \times 0 = 0$. So the value is 9.

Time check: about 20 seconds.

When it does not work: the expression may not contain the combination so neatly. Then divide it by the quadratic and keep only the remainder. For the same $x$,

$$x^3 - 5x^2 + 3 = (x^2 - 6x + 2)(x + 1) + 4x + 1$$

so its value is $4x + 1 = 13 + 4\sqrt{7}$. The remainder can still contain $x$, so you finish by putting the surd in.

The $a + b + c$ Family in One Line

When to use it: three letters, with some of $a + b + c$, $a^2 + b^2 + c^2$, $ab + bc + ca$, $abc$ and $a^3 + b^3 + c^3$ given.

Fast way: call $S = a + b + c$, $Q = a^2 + b^2 + c^2$ and $P = ab + bc + ca$. Two identities do all the work:

$$S^2 = Q + 2P$$

$$a^3 + b^3 + c^3 - 3abc = S(Q - P)$$

So any two of $S$, $Q$ and $P$ give the third, and then the cube expression. Also $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{P}{abc}$.

Question 9

If $a + b + c = 12$ and $a^2 + b^2 + c^2 = 56$, find $a^3 + b^3 + c^3 - 3abc$.

Answer. $P = \frac{144 - 56}{2} = 44$. Then $S(Q - P) = 12 \times (56 - 44) = 12 \times 12$ = 144. The numbers 2, 4 and 6 fit the data, and $8 + 64 + 216 - 3 \times 48 = 144$.

Time check: about 15 seconds.

When it does not work:

• The bracket is $Q - P$, not $Q - 2P$. It is easy to mix it up with $S^2 = Q + 2P$.
• If $a + b + c = 0$, the identity gives $a^3 + b^3 + c^3 = 3abc$ directly. The next method covers it.
• For $a^3 + b^3 + c^3$ alone you also need $abc$: $a^3 + b^3 + c^3 = 3abc + S(Q - P)$.

Spotting $a + b + c = 0$, Even in Shifted Brackets

When to use it: three cubes and a product, such as $(x - y)^3 + (y - z)^3 + (z - x)^3$, or brackets with numbers taken away, such as $(p - 3)^3 + (q - 7)^3 + (r - 5)^3$.

Fast way: add the three bases before you expand anything. If they add up to 0, the sum of the cubes is 3 times their product.

• $(x - y) + (y - z) + (z - x) = 0$, so $(x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x)$.
• If $p + q + r = 15$, then $(p - 3) + (q - 7) + (r - 5) = 15 - 15 = 0$. So $(p - 3)^3 + (q - 7)^3 + (r - 5)^3 = 3(p - 3)(q - 7)(r - 5)$.
• Two brackets and a number: if $a + b = k$, then $a + b + (-k) = 0$, so $a^3 + b^3 - k^3 = -3abk$, that is $a^3 + b^3 + 3kab = k^3$.

Question 10

If $p + q = 19$, find $(p - 7)^3 + (q - 5)^3 + 21(p - 7)(q - 5)$.

Answer. Let $a = p - 7$ and $b = q - 5$. Then $a + b = 19 - 12 = 7$, and $21 = 3 \times 7$. So the expression is $a^3 + b^3 + 3ab(a + b) = (a + b)^3 = 7^3$ = 343.

Time check: about 20 seconds.

When it does not work:

• The bases must add to exactly 0, or to the number that sits in the product term. Check the sum first, not after expanding.
• Signs stay in the product. With $a = 5$, $b = -2$, $c = -3$ (sum 0), $3abc = 3 \times 5 \times (-2) \times (-3) = 90$, and indeed $125 - 8 - 27 = 90$.
• In a fraction such as $\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{(x - y)(y - z)(z - x)} = 3$, the bottom must not be 0, so $x$, $y$ and $z$ are taken as different.

It also works for brackets of squares: $(x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2) = 0$.

Sum of Squares Equal to Zero

When to use it: one equation in two or three letters with squares and a number, such as $x^2 + y^2 + z^2 + 8x - 4y - 6z + 29 = 0$, and a value like $x + y + z$ is asked.

Fast way: complete the square for each letter on its own: half of the letter's coefficient, squared, is the number that letter needs. Check that these numbers add up to the constant. If they do, the equation is a sum of squares equal to 0, and for real numbers each square must be 0.

Question 11

If $x^2 + y^2 + z^2 + 8x - 4y - 6z + 29 = 0$, find $x + y + z$.

Answer. $x$ needs $4^2 = 16$, $y$ needs $2^2 = 4$ and $z$ needs $3^2 = 9$. These add up to 29, so the equation is $(x + 4)^2 + (y - 2)^2 + (z - 3)^2 = 0$. Each bracket is 0, so $x = -4$, $y = 2$, $z = 3$, and $x + y + z$ = 1.

Time check: about 20 seconds.

Equal to $ab + bc + ca$

$a^2 + b^2 + c^2 = ab + bc + ca$ is a sum of squares in disguise. Move everything to one side and double it: $(a - b)^2 + (b - c)^2 + (c - a)^2 = 0$. So $a = b = c$, and any expression is now quick, for instance $\frac{a + b}{c} = 2$.

When it does not work:

• The numbers do not add up to the constant. With $+20$ in place of $+29$, the equation becomes $(x + 4)^2 + (y - 2)^2 + (z - 3)^2 = 9$, which has endless solutions and no single answer.
• A number in front of a square. $4x^2 - 12x$ is $(2x - 3)^2 - 9$: write the letter's part as $(2x)^2$ first.
• It needs real numbers. SSC always means real numbers in these questions.

$8a^3 + b^3 + 27c^3 = 18abc$ and Its Cousins

When to use it: three cubes on one side and a multiple of $abc$ on the other, such as $8a^3 + b^3 + 27c^3 = 18abc$.

Fast way: write each term as a cube, here $(2a)^3 + b^3 + (3c)^3$, and check that the right side is exactly 3 times the product of the three bases: $3 \times 2a \times b \times 3c = 18abc$. Then the equation is $p^3 + q^3 + r^3 = 3pqr$, and

$$p^3 + q^3 + r^3 - 3pqr = \frac{1}{2}(p + q + r)\left[(p - q)^2 + (q - r)^2 + (r - p)^2\right] = 0$$

So either $p + q + r = 0$ or $p = q = r$. For positive numbers it must be $p = q = r$, so here $2a = b = 3c$.

Question 12

If $x$, $y$ and $z$ are positive and $64x^3 + y^3 + 27z^3 = 36xyz$, find $x : y : z$.

Answer. The bases are $4x$, $y$ and $3z$, and $3 \times 4 \times 1 \times 3 = 36$, so the shortcut applies. For positive values, $4x = y = 3z$. Take this common value as 12, the LCM of 4 and 3: then $x = 3$, $y = 12$, $z = 4$. So $x : y : z$ = 3 : 12 : 4.

Time check: about 20 seconds.

Two cousins

• $a + b = 2c$: then $(a - c) + (b - c) = 0$, so $(a - c)^3 + (b - c)^3 = 0$.
• $a + b = c$: then $a + b + (-c) = 0$, so $a^3 + b^3 - c^3 = -3abc$.

When it does not work:

• The number before $xyz$ must be exactly 3 times the product of the cube roots. With $64x^3 + y^3 + 27z^3 = 24xyz$ it is not, and the shortcut gives nothing.
• Without the word positive, $4x + y + 3z = 0$ is also possible. A question that says $4x + y + 3z \neq 0$ is pointing you to $4x = y = 3z$.
• The ratio is not the list of cube roots. $4x = y = 3z$ gives 3 : 12 : 4, not 4 : 1 : 3.

Remainder Theorem in One Substitution

When to use it: the remainder when a polynomial is divided by $x - a$, $x + a$ or $ax - b$; the value of $k$ that makes something a factor; or the number to take away so that a factor appears.

Fast way: never do the long division. The remainder of $f(x) \div (x - a)$ is $f(a)$.

• Divisor $x + 2$: put $x = -2$.
• Divisor $2x - 1$: put $x = \frac{1}{2}$. For $2x^3 - 5x^2 + 4x - 7$ this gives $\frac{1}{4} - \frac{5}{4} + 2 - 7 = -6$.
• Factor: set the value equal to 0 and solve for $k$.
• The number to subtract so that $x - a$ becomes a factor is the remainder $f(a)$ itself.

Question 13

Find $k$ if $x + 2$ is a factor of $x^3 + kx^2 - 4x + 12$.

Answer. Put $x = -2$: $-8 + 4k + 8 + 12 = 0$, so $4k = -12$ and $k$ = $-3$. Check: $x^3 - 3x^2 - 4x + 12 = (x - 3)(x - 2)(x + 2)$.

Time check: about 15 seconds.

[Tier 2] Dividing by a quadratic

If the divisor is $(x - a)(x - b)$, the remainder is a line $px + q$, with $f(a) = pa + q$ and $f(b) = pb + q$. Two values give two equations for $p$ and $q$.

When it does not work:

• The sign. $x + 2$ means $x = -2$. Putting $x = 2$ is the usual slip.
• For $ax - b$, put $x = \frac{b}{a}$, not $x = b$.
• The remainder theorem gives the remainder, not the quotient. If the quotient is asked, divide or factorise properly.

Quadratic Roots Without Solving

When to use it: $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$, and you need $\alpha^2 + \beta^2$, $\frac{1}{\alpha} + \frac{1}{\beta}$, $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ or a new equation.

Fast way: $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. These are the $s$ and $p$ from the first method in this section, so the same table works.

• Equal roots: $b^2 = 4ac$. For $kx^2 - 12x + 9 = 0$, $144 = 36k$, so $k = 4$.
• One root given: put it in to find $k$, or get the other root from the product.
• Roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$: swap the end coefficients, $cx^2 + bx + a = 0$.
• Any new pair of roots: $x^2 - (\text{sum})x + \text{product} = 0$.

Question 14

If $\alpha$ and $\beta$ are the roots of $3x^2 - 8x + 2 = 0$, find $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.

Answer. $s = \frac{8}{3}$ and $p = \frac{2}{3}$. Then $\alpha^2 + \beta^2 = \frac{64}{9} - \frac{4}{3} = \frac{52}{9}$, and dividing by $p$, $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{52}{9} \times \frac{3}{2}$ = $\frac{26}{3}$.

Time check: about 20 seconds.

When it does not work:

• The sum is $-\frac{b}{a}$. Taking $+\frac{8}{3}$ as the sum for $3x^2 + 8x + 2 = 0$ is a common slip.
• Bring every term to one side first. $x^2 = 4x - 1$ is $x^2 - 4x + 1 = 0$, so its sum of roots is 4.

[Tier 2] A cubic $ax^3 + bx^2 + cx + d = 0$ has $\alpha + \beta + \gamma = -\frac{b}{a}$, $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$ and $\alpha\beta\gamma = -\frac{d}{a}$.