Three-Variable Identities and Special Conditions

The Square of Three Terms

Treat $a + b$ as one term and use the two-letter square:

$$(a + b + c)^2 = (a + b)^2 + 2(a + b)c + c^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$

In words: every term squared once, and every pair multiplied and doubled.

Signs. The sign of each pair is the product of the two signs. In $(a - b + c)^2$ the pairs are $a \times (-b)$, $(-b) \times c$ and $c \times a$, so

$$(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca$$

With numbers in front, square each term fully and double each pair product:

$$(2x - y - 3z)^2 = 4x^2 + y^2 + 9z^2 - 4xy + 6yz - 12zx$$

The $yz$ term is positive because $(-y) \times (-3z)$ is positive.

Three building blocks. The identity links the sum $a + b + c$, the sum of squares $a^2 + b^2 + c^2$, and the pair sum $ab + bc + ca$. Any two give the third:

$$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$$

$$ab + bc + ca = \frac{(a + b + c)^2 - (a^2 + b^2 + c^2)}{2}$$

• $a + b + c = 9$ and $ab + bc + ca = 20$: $a^2 + b^2 + c^2 = 81 - 40 = 41$. (1, 2 and 6 fit.)
• $a + b + c = 15$ and $a^2 + b^2 + c^2 = 93$: $ab + bc + ca = \frac{225 - 93}{2} = 66$.
• $a^2 + b^2 + c^2 = 53$ and $ab + bc + ca = 34$: $(a + b + c)^2 = 53 + 68 = 121$, so $a + b + c = 11$ if the numbers are positive (and $\pm 11$ otherwise).

A check for real numbers. For real numbers, $a^2 + b^2 + c^2$ is never smaller than $ab + bc + ca$; the reason comes later in this section. So data like $a^2 + b^2 + c^2 = 20$ with $ab + bc + ca = 25$ cannot come from real numbers.

A cuboid application. For a cuboid with edges $l$, $b$ and $h$, the 12 edges add to $4(l + b + h)$, the total surface area is $2(lb + bh + hl)$, and the longest diagonal is $\sqrt{l^2 + b^2 + h^2}$. If the edges add to 68 cm and the surface area is 168 sq cm, then $l + b + h = 17$ and $l^2 + b^2 + h^2 = 289 - 168 = 121$, so the diagonal is 11 cm. (The cuboid is 2 cm by 6 cm by 9 cm.)

Symmetric Expressions: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ and Friends

An expression that stays the same when you swap any two letters is called symmetric. Every symmetric expression SSC asks about can be written with three building blocks: $a + b + c$, $ab + bc + ca$ and $abc$. Put everything over a common bottom and the building blocks appear.

$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ca + ab}{abc}$$

Each top is the product of the two letters that are missing from that bottom: for $\frac{1}{a}$ the top becomes $bc$.

Expression	Over the common bottom $abc$
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$	$\frac{ab + bc + ca}{abc}$
$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$	$\frac{a + b + c}{abc}$
$\frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab}$	$\frac{a^2 + b^2 + c^2}{abc}$

Try them on $a = 2$, $b = 3$ and $c = 6$, where $a + b + c = 11$, $ab + bc + ca = 36$ and $abc = 36$:

• $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{36}{36} = 1$, and indeed $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$.
• $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{11}{36}$.
• $\frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab} = \frac{49}{36}$, because $a^2 + b^2 + c^2 = 121 - 72 = 49$.

[Tier 2] Two longer ones.

$$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2 - \frac{2(a + b + c)}{abc}$$

This is the three-term square used on $\frac{1}{a}$, $\frac{1}{b}$ and $\frac{1}{c}$. Their pair products are $\frac{1}{ab}$, $\frac{1}{bc}$ and $\frac{1}{ca}$, which add to $\frac{a + b + c}{abc}$. For 2, 3 and 6 it gives $1 - \frac{22}{36} = \frac{7}{18}$, and $\frac{1}{4} + \frac{1}{9} + \frac{1}{36}$ is $\frac{7}{18}$ too.

$$(a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - abc$$

For 2, 3 and 6: $11 \times 36 - 36 = 360$, and $5 \times 9 \times 8 = 360$.

When the reciprocals add to zero. If $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$ (none of the letters being zero), the top $ab + bc + ca$ must be 0. The three-term square then shrinks to $(a + b + c)^2 = a^2 + b^2 + c^2$. SSC asks this as "find $(a + b + c)^2$", and puts $a^2 + b^2 + c^2$ in the options to catch you.

$a^3 + b^3 + c^3 - 3abc$ in Two Forms

This is the main three-letter identity:

$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$

You can check it by multiplying out the right side. The cross terms such as $a^2b$ and $ab^2$ cancel in pairs, the three $abc$ terms collect into $-3abc$, and the cubes are left.

The long bracket in two other forms.

$$a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]$$

Expand the three squares: each of $a^2$, $b^2$ and $c^2$ appears twice, and each pair product appears once with $-2$, so the total is exactly twice the bracket. That gives the half-sum-of-squares form:

$$a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c)\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]$$

The bracket also comes straight from the building blocks:

$$a^2 + b^2 + c^2 - ab - bc - ca = (a + b + c)^2 - 3(ab + bc + ca)$$

because $(a + b + c)^2$ holds $a^2 + b^2 + c^2$ plus $2(ab + bc + ca)$, and taking away $3(ab + bc + ca)$ leaves one lot with a minus sign.

The sum of cubes from the building blocks. Rearranging the identity,

$$a^3 + b^3 + c^3 = 3abc + (a + b + c)\left[(a + b + c)^2 - 3(ab + bc + ca)\right]$$

With $a + b + c = 9$, $ab + bc + ca = 24$ and $abc = 20$: $a^3 + b^3 + c^3 = 60 + 9 \times (81 - 72) = 60 + 81 = 141$. (The numbers are 2, 2 and 5, and $8 + 8 + 125 = 141$.)

The long bracket at the bottom. If a question divides by the long bracket, the answer is the sum:

$$\frac{a^3 + b^3 + c^3 - 3abc}{a^2 + b^2 + c^2 - ab - bc - ca} = a + b + c$$

If it divides by the sum of the three squared differences instead, the answer is half the sum, $\frac{a + b + c}{2}$. Both need $a$, $b$ and $c$ not all equal, or the bottom is zero.

[SSC Tip] The long bracket is $(a + b + c)^2$ minus three times the pair sum, while $a^2 + b^2 + c^2$ is $(a + b + c)^2$ minus two times the pair sum. Mixing up the 2 and the 3 is the classic slip in these questions.

Working Back, and Differences in Place of Values

From the sum and the whole expression. "If $a + b + c = 11$ and $a^3 + b^3 + c^3 - 3abc = 110$, find $ab + bc + ca$ and $a^2 + b^2 + c^2$."

1. Divide by the sum: the long bracket is $110 \div 11 = 10$, so $a^2 + b^2 + c^2 - (ab + bc + ca) = 10$.
2. The three-term square gives $a^2 + b^2 + c^2 + 2(ab + bc + ca) = 121$.
3. Take step 1 from step 2: $3(ab + bc + ca) = 111$, so $ab + bc + ca = 37$ and $a^2 + b^2 + c^2 = 37 + 10 = 47$.

In one line: $ab + bc + ca = \frac{(a + b + c)^2 - \text{bracket}}{3} = \frac{121 - 10}{3} = 37$.

From the sum and the sum of squares. If $a + b + c = 5$ and $a^2 + b^2 + c^2 = 13$, then $ab + bc + ca = \frac{25 - 13}{2} = 6$, the bracket is $13 - 6 = 7$, and $a^3 + b^3 + c^3 - 3abc = 5 \times 7 = 35$. If you are also told $abc = 0$, then $a^3 + b^3 + c^3 = 35$ itself; the numbers 0, 2 and 3 fit.

Differences given instead of values. Often the question gives $a - b$ and $b - c$. The three differences always add to zero:

$$(a - b) + (b - c) + (c - a) = 0$$

So if $a - b = 3$ and $b - c = 2$, then $c - a = -5$. Now use the half-sum-of-squares form:

$$a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}(9 + 4 + 25) = 19$$

The sign of $c - a$ does not matter once it is squared, but its size does: the third gap is the two given gaps added, 5, not their difference.

Letters in a row. If $a = x + 2$, $b = x + 4$ and $c = x + 6$, the gaps are 2, 2 and 4, so the bracket is $\frac{1}{2}(4 + 4 + 16) = 12$. The sum is $3x + 12$. So

$$a^3 + b^3 + c^3 - 3abc = 12(3x + 12) = 36(x + 4)$$

No cubing of $x + 2$ needed.

[SSC Tip] With two gaps given, the third gap is their sum with a minus sign. Squaring only the two given gaps gives $\frac{1}{2}(9 + 4) = 6.5$ in the example above, and SSC likes to keep that number as an option. Write all three gaps down before you square anything.

When $a + b + c = 0$

Put $a + b + c = 0$ into the identity. The right side is 0 times a bracket, so it is 0:

$$a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc$$

You can also see it directly. If $c = -(a + b)$, then

$$c^3 = -(a + b)^3 = -a^3 - b^3 - 3ab(a + b) = -a^3 - b^3 + 3abc$$

because $-(a + b)$ is $c$. Move $a^3$ and $b^3$ across, and $a^3 + b^3 + c^3 = 3abc$ appears.

This one fact turns large cube expressions into a single product.

• If $a + b + c = 0$ and $abc = 4$, then $a^3 + b^3 + c^3 = 12$.
• The three differences always add to zero, so

$$(x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x)$$

With $x = 4$, $y = 1$ and $z = 6$, the brackets are 3, $-5$ and 2, and $27 - 125 + 8 = -90$, which is $3 \times 3 \times (-5) \times 2$.

• With coefficients the brackets still cancel: $(2a - 3b) + (3b - c) + (c - 2a) = 0$, so

$$(2a - 3b)^3 + (3b - c)^3 + (c - 2a)^3 = 3(2a - 3b)(3b - c)(c - 2a)$$

[Tier 2] Squares of differences. The brackets $a^2 - b^2$, $b^2 - c^2$ and $c^2 - a^2$ also add to zero. So the top and bottom of this fraction are both "3 times the product":

$$\frac{(a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3}{(a - b)^3 + (b - c)^3 + (c - a)^3} = \frac{3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2)}{3(a - b)(b - c)(c - a)} = (a + b)(b + c)(c + a)$$

since $a^2 - b^2 = (a - b)(a + b)$, and the same for the other two.

The reverse needs care. $a^3 + b^3 + c^3 = 3abc$ does not force $a + b + c = 0$. It also holds when $a = b = c$, since then both sides are $3a^3$. That second case is one of the special conditions later in this section.

[SSC Tip] Before expanding any sum of three cubes, add the three bases with their signs. If they add to zero, write 3 times their product and stop. Keep the 3: in $\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{(x - y)(y - z)(z - x)}$ the answer is 3, not 1.

Spotting Three Brackets That Add to Zero

The skill here is noticing that the brackets cancel. Add them with their signs and see whether every letter disappears.

Brackets	Their sum
$x - y$, $y - z$, $z - x$	0
$2x - 3y$, $3y - 4z$, $4z - 2x$	0
$x + y - 2z$, $y + z - 2x$, $z + x - 2y$	0
$p - 1$, $q - 2$, $r - 3$ when $p + q + r = 6$	0

Three brackets and a condition. "If $p + q + r = 6$, find $(p - 1)^3 + (q - 2)^3 + (r - 3)^3 - 3(p - 1)(q - 2)(r - 3)$." The brackets add to $p + q + r - 6 = 0$, so the whole expression is 0.

Two brackets and a number. "If $x + y = 15$, find $(x - 8)^3 + (y - 3)^3 + 12(x - 8)(y - 3)$." Call the brackets $A = x - 8$ and $B = y - 3$. They add to $x + y - 11 = 4$. So $A$, $B$ and $-4$ add to zero, and the cube fact gives

$$A^3 + B^3 + (-4)^3 = 3AB(-4) \Rightarrow A^3 + B^3 + 12AB = 64$$

The answer is 64 for every $x$ and $y$ with $x + y = 15$. The clue is the 12 in front of the product: it is $3 \times 4$, three times what the two brackets add up to. In two-letter language this is just $A + B = 4 \Rightarrow A^3 + B^3 + 3 \times 4 \times AB = 4^3$, the cube rule from the Square and Cube Identities section.

[Tier 2] Cube roots that add to zero. If $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$, call the cube roots $u$, $v$ and $w$. They add to zero, so $u^3 + v^3 + w^3 = 3uvw$, which says $a + b + c = 3\sqrt[3]{abc}$. Cubing both sides,

$$(a + b + c)^3 = 27abc$$

Check with $u = 1$, $v = 1$ and $w = -2$: then $a = 1$, $b = 1$ and $c = -8$, so $a + b + c = -6$, and $(-6)^3 = -216 = 27 \times (-8)$.

How the question tries to hide it. The brackets may be written in a different order, with the signs moved outside, as in $(y - x)^3$ in place of $(x - y)^3$, or with one bracket replaced by a number. In each case, rewrite until you have three bases and check their sum.

More Results When $a + b + c = 0$

When $a + b + c = 0$, every pair adds to minus the third letter:

$$a + b = -c \qquad b + c = -a \qquad c + a = -b$$

Swap these in and many expressions collapse.

• $(a + b)(b + c)(c + a) = (-c)(-a)(-b) = -abc$.
• $\frac{a + b}{c} + \frac{b + c}{a} + \frac{c + a}{b} = -1 - 1 - 1 = -3$.
• $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = \frac{a^3 + b^3 + c^3}{abc} = \frac{3abc}{abc} = 3$.
• Squaring $a + b + c = 0$ gives $a^2 + b^2 + c^2 + 2(ab + bc + ca) = 0$, so $a^2 + b^2 + c^2 = -2(ab + bc + ca)$.

The $a^2 - bc$ family. Put $a = -(b + c)$: $a^2 - bc = b^2 + 2bc + c^2 - bc = b^2 + bc + c^2$. The other two work out the same, so

$$a^2 - bc = b^2 - ca = c^2 - ab$$

A quick proof of the first equals sign: $(a^2 - bc) - (b^2 - ca) = (a - b)(a + b) + c(a - b) = (a - b)(a + b + c) = 0$. Also $a^2 + b^2 + c^2 = (b + c)^2 + b^2 + c^2 = 2(b^2 + bc + c^2)$, so

$$\frac{a^2 + b^2 + c^2}{a^2 - bc} = 2$$

[Tier 2] Two more.

• $a^2 + b^2 - c^2 = a^2 + b^2 - (a + b)^2 = -2ab$. So $\frac{1}{a^2 + b^2 - c^2} + \frac{1}{b^2 + c^2 - a^2} + \frac{1}{c^2 + a^2 - b^2} = -\frac{1}{2}\left(\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}\right) = -\frac{a + b + c}{2abc} = 0$.
• $a^4 + b^4 + c^4 = \frac{(a^2 + b^2 + c^2)^2}{2}$, which is the same as $2(ab + bc + ca)^2$.

Test with numbers. Any three numbers that add to zero will do, and $a = 1$, $b = 1$, $c = -2$ is the easiest set. For $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$ it gives $\frac{1}{-2} + \frac{1}{-2} + \frac{4}{1} = 3$. For $a^4 + b^4 + c^4$ it gives $1 + 1 + 16 = 18$, and $\frac{(1 + 1 + 4)^2}{2} = 18$.

When an unfamiliar expression comes with the condition $a + b + c = 0$, do not hunt for an identity. Put in 1, 1 and $-2$ (or 1, 2 and $-3$ if a bottom would become zero) and compare with the options. It takes a few seconds and it is always allowed, because the answer has to hold for every such triple.

The Cases $a + b = c$ and $a + b = 2c$

Both conditions are a sum equal to zero in disguise. Move everything to one side and choose the three brackets.

$a + b = c$. Then $a + b + (-c) = 0$. Use the cube fact on $a$, $b$ and $-c$:

$$a^3 + b^3 + (-c)^3 = 3ab(-c) \Rightarrow a^3 + b^3 - c^3 = -3abc$$

So $a^3 + b^3 - c^3 + 3abc = 0$, or written the other way round, $a^3 + b^3 + 3abc = c^3$. Check with $a = 1$, $b = 2$, $c = 3$: $1 + 8 + 18 = 27$.

The same works for any arrangement. If $x = y + z$, then $x - y - z = 0$, the three bases are $x$, $-y$ and $-z$, and $x^3 - y^3 - z^3 = 3x(-y)(-z) = 3xyz$.

$a + b = 2c$. This says $c$ is exactly halfway between $a$ and $b$ (the three numbers are in arithmetic progression). Split the $2c$ as $c + c$:

$$(a - c) + (b - c) = 0 \qquad \text{so} \qquad a - c = -(b - c) = c - b$$

The two gaps from $c$ are the same size with opposite signs. Everything follows from this one line.

• $(a - c)^3 + (b - c)^3 = 0$, since one cube is the negative of the other.
• $\frac{1}{a - c} + \frac{1}{b - c} = 0$.
• $\frac{a}{a - c} + \frac{b}{b - c} = \frac{a}{a - c} - \frac{b}{a - c} = \frac{a - b}{a - c}$, and since $a - c = \frac{a - b}{2}$, this is 2.

Check with $a = 7$, $b = 3$ and $c = 5$: $2^3 + (-2)^3 = 0$, and $\frac{7}{2} + \frac{3}{-2} = 2$.

Which one is it? If the condition has $c$ alone on one side, think "$a + b + (-c) = 0$" and use the cube fact. If it has $2c$, think "$c$ is the middle value, and the two gaps cancel".

Sum of Squares Equal to Zero

For real numbers a square is never negative. So if several squares add up to zero, each square must be zero. Two kinds of SSC question rest on this.

Kind 1: $a^2 + b^2 + c^2 = ab + bc + ca$. Move everything to one side and use the half-sum-of-squares form:

$$a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right] = 0$$

So $a - b$, $b - c$ and $c - a$ are all zero, which means $a = b = c$. After that every expression is easy: $\frac{(a + b + c)^2}{a^2 + b^2 + c^2} = \frac{9a^2}{3a^2} = 3$. This also explains the rule from the start of the section: for real numbers $a^2 + b^2 + c^2 \ge ab + bc + ca$, with equality only when all three are equal.

The scaled version. "If $4a^2 + b^2 + c^2 = 2ab + bc + 2ca$, find $a : b : c$." Write $X = 2a$, $Y = b$ and $Z = c$. Then $4a^2 = X^2$, $2ab = XY$, $bc = YZ$ and $2ca = ZX$, so the condition is $X^2 + Y^2 + Z^2 = XY + YZ + ZX$, and $X = Y = Z$. That gives $2a = b = c$, so $a : b : c = 1 : 2 : 2$.

Kind 2: completing squares. "$x^2 + y^2 + z^2 - 4x + 6y - 10z + 38 = 0$. Find $x$, $y$ and $z$." Keep each letter with its own term and complete the square:

$$(x^2 - 4x + 4) + (y^2 + 6y + 9) + (z^2 - 10z + 25) = 0$$

The three constants 4, 9 and 25 add to exactly 38, which tells you the question is built this way. So $(x - 2)^2 + (y + 3)^2 + (z - 5)^2 = 0$, and $x = 2$, $y = -3$, $z = 5$.

To find each constant, halve the coefficient of the letter and square it: $-4x$ gives $(-2)^2 = 4$, $+6y$ gives $3^2 = 9$, and so on. The sign inside the bracket follows the sign of that coefficient.

With two letters it is the same: $p^2 + q^2 + 2 = 2(p + q)$ rearranges to $(p - 1)^2 + (q - 1)^2 = 0$, so $p = q = 1$.

[Tier 2] A mixed term. $x^2 + 2y^2 - 2xy - 4y + 4 = 0$ has an $xy$ term. Split $2y^2$ into $y^2 + y^2$ and pair one of them with the $x$ terms: $(x^2 - 2xy + y^2) + (y^2 - 4y + 4) = (x - y)^2 + (y - 2)^2 = 0$. So $y = 2$, and $x = y = 2$.

At the end, read the signs carefully: $(y + 3)^2 = 0$ means $y = -3$, and the wrong-sign value is a favourite option.

When $a^3 + b^3 + c^3 = 3abc$

From the half-sum-of-squares form, $a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c)\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]$. If the left side is zero, one of the two factors must be zero:

$$a^3 + b^3 + c^3 = 3abc \Rightarrow a + b + c = 0 \ \text{ or } \ a = b = c$$

• If the question says $a + b + c \neq 0$, then $a = b = c$.
• If it says $a$, $b$ and $c$ are positive, their sum cannot be zero, so again $a = b = c$.
• If it says neither, both cases are possible, and an option like "$a + b + c = 0$" may be the one wanted.

With $a = b = c$: $\frac{a^2 + b^2 + c^2}{ab + bc + ca} = 1$, $\frac{a + b}{c} = 2$ and $\frac{(a + b + c)^3}{abc} = 27$.

The scaled version. "$27x^3 + 8y^3 + z^3 = 18xyz$, where $x$, $y$ and $z$ are positive. Find $x : y : z$."

1. Take cube roots of the coefficients: $27x^3 = (3x)^3$, $8y^3 = (2y)^3$ and $z^3$ stays as it is.
2. Check the right side: $3 \times 3x \times 2y \times z = 18xyz$. It matches, so this is $A^3 + B^3 + C^3 = 3ABC$ with $A = 3x$, $B = 2y$ and $C = z$.
3. Positive values rule out $A + B + C = 0$, so $3x = 2y = z$.
4. Give the common value a number that every multiplier divides, here 6. Then $x = 2$, $y = 3$ and $z = 6$, so $x : y : z = 2 : 3 : 6$.

If the number on the right is not 3 times the product of the cube roots, the trick does not apply.

Spotting it fast. Coefficients 1, 8, 27, 64 or 125 on the cubes are the signal. Multiply their cube roots, multiply by 3, and compare with the number in front of the product.

Condition	Cube roots	Result for positive values
$27x^3 + 8y^3 + z^3 = 18xyz$	3, 2, 1	$3x = 2y = z$
$x^3 + 8y^3 + 64z^3 = 24xyz$	1, 2, 4	$x = 2y = 4z$
$a^3 + 64b^3 + 125c^3 = 60abc$	1, 4, 5	$a = 4b = 5c$

Without the word "positive", the first row also allows $3x + 2y + z = 0$.

Solved Examples

Question 1

If $a + b + c = 16$ and $ab + bc + ca = 79$, what is $a^2 + b^2 + c^2$? (a) 177 (b) 98 (c) 414 (d) 19

Answer. Use $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 256 - 158$, which gives (b) 98. (3, 5 and 8 fit: $9 + 25 + 64 = 98$.)

Question 2

If $a^2 + b^2 + c^2 = 78$ and $ab + bc + ca = 59$, where $a$, $b$ and $c$ are positive, what is $a + b + c$? (a) $\sqrt{137}$ (b) $196$ (c) $\sqrt{19}$ (d) $14$

Answer. $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 78 + 118 = 196$. The numbers are positive, so their sum is positive: (d) $14$. (2, 5 and 7 fit.) 196 is still a square; its root is the answer.

Question 3

Which of these is the expansion of $(x - 2y - 3z)^2$? (a) $x^2 + 4y^2 + 9z^2 - 4xy + 12yz - 6zx$ (b) $x^2 + 4y^2 + 9z^2 - 4xy - 12yz - 6zx$ (c) $x^2 + 4y^2 + 9z^2 + 4xy + 12yz + 6zx$ (d) $x^2 + 4y^2 + 9z^2 - 2xy + 6yz - 3zx$

Answer. Square each term: $x^2$, $4y^2$, $9z^2$. Then each pair, doubled, with the product of its signs: $2 \times x \times (-2y) = -4xy$, $2 \times (-2y) \times (-3z) = +12yz$ and $2 \times (-3z) \times x = -6zx$. That is (a) $x^2 + 4y^2 + 9z^2 - 4xy + 12yz - 6zx$. The $yz$ term is positive because two minus signs multiply; option (b) misses that.

Question 4

If $a + b + c = 17$, $ab + bc + ca = 66$ and $abc = 72$, what is $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$? (a) $\frac{17}{72}$ (b) $\frac{12}{11}$ (c) $\frac{11}{12}$ (d) $\frac{66}{17}$

Answer. Put the three fractions over $abc$: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ca + ab}{abc} = \frac{66}{72}$, which is (c) $\frac{11}{12}$. (The numbers are 2, 3 and 12, and $\frac{1}{2} + \frac{1}{3} + \frac{1}{12} = \frac{11}{12}$.) Putting the sum on top gives $\frac{17}{72}$, which answers a different question, $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$.

Question 5

If $a + b + c = 10$, $a^2 + b^2 + c^2 = 46$ and $abc = 18$, what is $\frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab}$? (a) $\frac{23}{9}$ (b) $\frac{5}{9}$ (c) $\frac{3}{2}$ (d) $\frac{9}{23}$

Answer. Over the common bottom $abc$, the tops become $a^2$, $b^2$ and $c^2$: $\frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab} = \frac{a^2 + b^2 + c^2}{abc} = \frac{46}{18}$, which is (a) $\frac{23}{9}$. (The numbers are 1, 3 and 6.)

Question 6

If $a + b + c = 11$, $ab + bc + ca = 31$ and $abc = 21$, what is $a^3 + b^3 + c^3$? (a) 308 (b) 371 (c) 712 (d) 1331

Answer. The long bracket is $(a + b + c)^2 - 3(ab + bc + ca) = 121 - 93 = 28$. So $a^3 + b^3 + c^3 - 3abc = 11 \times 28 = 308$, and adding back $3abc = 63$ gives (b) 371. Check with 1, 3 and 7: $1 + 27 + 343 = 371$.