Square and Cube Identities

The Square Identities

An identity is true for every value of the letters. An equation such as $2x + 1 = 7$ is true only for $x = 3$, but $(a + b)^2 = a^2 + 2ab + b^2$ works for any $a$ and $b$ you choose. That is why you can always test an identity by putting in numbers.

Where it comes from. $(a + b)^2$ means $(a + b)(a + b)$. Multiply each term of the first bracket by each term of the second: $a \times a + a \times b + b \times a + b \times b = a^2 + 2ab + b^2$. Every identity below is found by the same kind of multiplying.

$$(a + b)^2 = a^2 + 2ab + b^2 \qquad (a - b)^2 = a^2 - 2ab + b^2$$

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

$$(a + b)^2 + (a - b)^2 = 2(a^2 + b^2) \qquad (a + b)^2 - (a - b)^2 = 4ab$$

$$(x + a)(x + b) = x^2 + (a + b)x + ab$$

When the terms carry numbers. In $(3x - 4y)^2$, the first term is $3x$ and the second is $4y$. Square each one fully and double their product:

$$(3x - 4y)^2 = 9x^2 - 2 \times 3x \times 4y + 16y^2 = 9x^2 - 24xy + 16y^2$$

Adding and subtracting the two squares. Adding $(a + b)^2$ and $(a - b)^2$ cancels the $2ab$ terms. Subtracting them cancels the squares. So $(2x + 5y)^2 + (2x - 5y)^2 = 2(4x^2 + 25y^2) = 8x^2 + 50y^2$, and $(2x + 5y)^2 - (2x - 5y)^2 = 4 \times 2x \times 5y = 40xy$. Two more forms of the same fact link the sum and the difference: $(a + b)^2 = (a - b)^2 + 4ab$ and $(a - b)^2 = (a + b)^2 - 4ab$.

$(x + a)(x + b)$ with signs. Add the two numbers for the middle coefficient and multiply them for the last term. $(x + 6)(x - 2) = x^2 + 4x - 12$, because $6 - 2 = 4$ and $6 \times (-2) = -12$. Likewise $(x - 5)(x - 3) = x^2 - 8x + 15$.

Perfect squares and the missing term. $16x^2 + kx + 25$ is a perfect square when the middle term is $2 \times 4x \times 5 = 40x$, so $k = 40$ (or $-40$ for $(4x - 5)^2$). To complete $x^2 + 10x$, take half of 10, which is 5, and add its square: $x^2 + 10x + 25 = (x + 5)^2$.

The same identities also make number work quick, such as $67^2 - 33^2$ or a big $(a + b)^2 - (a - b)^2$ over $ab$ (see Chapter 2).

Never write $(a + b)^2$ as $a^2 + b^2$. To see why, put $a = 1$ and $b = 1$: the left side is 4, the right side is 2. Expect an option that drops the $2ab$ term, or squares $x$ but not the 3 in $3x$, among the four choices.

Sum, Product and Difference: Getting One from the Others

A typical two-letter question gives you two of $a + b$, $a - b$, $ab$ and $a^2 + b^2$, and asks for something else. You never need $a$ and $b$ themselves. Rearrange the square identities:

$$a^2 + b^2 = (a + b)^2 - 2ab = (a - b)^2 + 2ab$$

$$(a - b)^2 = (a + b)^2 - 4ab \qquad (a + b)^2 = (a - b)^2 + 4ab$$

$$ab = \frac{(a + b)^2 - (a^2 + b^2)}{2} = \frac{(a + b)^2 - (a - b)^2}{4}$$

One pair, many answers. Say $a + b = 13$ and $ab = 36$.

Asked	Working	Value
$a^2 + b^2$	$169 - 2 \times 36$	97
$(a - b)^2$	$169 - 4 \times 36$	25
$a - b$ (for $a > b$)	$\sqrt{25}$	5
$a^2 - b^2$	$(a + b)(a - b) = 13 \times 5$	65
$\frac{a}{b} + \frac{b}{a}$	$\frac{a^2 + b^2}{ab}$	$\frac{97}{36}$
$\frac{1}{a} + \frac{1}{b}$	$\frac{a + b}{ab}$	$\frac{13}{36}$

The numbers are 9 and 4, and $81 + 16 = 97$ confirms the first line. In an exam you would not stop to find them.

Starting from $a^2 + b^2$ and $ab$. If $a^2 + b^2 = 113$ and $ab = 56$, with both positive, then $(a + b)^2 = 113 + 112 = 225$, so $a + b = 15$, and $(a - b)^2 = 113 - 112 = 1$, so $a - b = 1$ when $a > b$.

Starting from $a + b$ and $a^2 + b^2$. If $a + b = 12$ and $a^2 + b^2 = 90$, then $2ab = 144 - 90 = 54$ and $ab = 27$. (The numbers are 9 and 3.)

Do real values exist? For real numbers $(a - b)^2$ can never be negative, so $(a + b)^2$ must be at least $4ab$. Suppose a question gives $a + b = 4$ and $ab = 5$. The working for $a^2 + b^2 = 16 - 10 = 6$ still runs, but $(a - b)^2 = 16 - 20 = -4$, so there are no real $a$ and $b$ at all. If a question asks for $a - b$ from data like this, you have probably misread a number.

[SSC Tip] The sum of squares needs $2ab$, the square of the difference needs $4ab$. Most wrong options in these questions are what you get by swapping the two. And $a - b$ comes out of a square root, so it is $\pm$; take the positive value only when the question tells you $a > b$.

Fourth Powers: $a^4 + b^4$, $a^4 - b^4$ and $a^4 + a^2b^2 + b^4$

$a^4 + b^4$. Think of $a^2$ and $b^2$ as the two terms. Then $(a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4$, so

$$a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2$$

It is a two-step job: find $a^2 + b^2$ first, then square it and take away $2(ab)^2$. With $a - b = 3$ and $ab = 2$: $a^2 + b^2 = 9 + 4 = 13$, and $a^4 + b^4 = 169 - 2 \times 4 = 161$. The subtraction is $2a^2b^2 = 8$, not $2ab = 4$. If $s = a + b$ and $p = ab$, the whole thing in one line is $a^4 + b^4 = (s^2 - 2p)^2 - 2p^2$.

$a^4 - b^4$. Use the difference of squares twice:

$$a^4 - b^4 = (a^2 + b^2)(a^2 - b^2) = (a^2 + b^2)(a + b)(a - b)$$

So $x^4 - 81 = (x^2 + 9)(x + 3)(x - 3)$. The bracket $x^2 + 9$ is a sum of squares and does not break further with real numbers.

$a^4 + a^2b^2 + b^4$. Add and take away $a^2b^2$ so that a full square appears:

$$a^4 + a^2b^2 + b^4 = (a^4 + 2a^2b^2 + b^4) - a^2b^2 = (a^2 + b^2)^2 - (ab)^2$$

That is a difference of squares, so

$$a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)$$

With $b = 1$ it reads $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$, which also turns up on its own.

The usual question. "If $a^4 + a^2b^2 + b^4 = 21$ and $a^2 + ab + b^2 = 7$, find $a^2 - ab + b^2$, $a^2 + b^2$ and $ab$."

1. Divide: $a^2 - ab + b^2 = 21 \div 7 = 3$.
2. Add the two brackets: $2(a^2 + b^2) = 7 + 3 = 10$, so $a^2 + b^2 = 5$.
3. Take one from the other: $2ab = 7 - 3 = 4$, so $ab = 2$.

From here $(a + b)^2 = 5 + 4 = 9$, and so on. (The numbers are 2 and 1: $16 + 4 + 1 = 21$.)

The Cube Identities

Multiply $(a + b)$ by $(a + b)^2 = a^2 + 2ab + b^2$ and collect the terms:

$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a + b)$$

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

The sum and difference of two cubes each have two useful forms:

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

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

Two forms, two jobs. The bracket forms are for factorising and cancelling. The forms with $3ab$ are for working out a value from a sum, a difference and a product, which is covered next.

Signs in the bracket form. $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$. The first bracket has the same sign as the question, the middle term of the second bracket has the opposite sign, and the last term is always positive. Some students remember this as SOAP.

Coefficients 1, 3, 3, 1. For $(3x - 2y)^3$ the four terms are $(3x)^3$, $3(3x)^2(2y)$, $3(3x)(2y)^2$ and $(2y)^3$, with signs plus, minus, plus, minus:

$$(3x - 2y)^3 = 27x^3 - 54x^2y + 36xy^2 - 8y^3$$

Spotting a full cube. In $x^3 + 6x^2 + 12x + 8$, the first term is $x^3$ and the last is $8 = 2^3$. Check the middle: $3 \times x^2 \times 2 = 6x^2$ and $3 \times x \times 4 = 12x$. Both match, so it is $(x + 2)^3$, and its value at $x = 3$ is $5^3 = 125$ without any long working. In the same way $8x^3 - 12x^2 + 6x - 1 = (2x - 1)^3$; the alternating signs tell you it is a minus cube.

The $3ab$ term when the sum is a number. If $x + y = 2$, then $(x + y)^3 = x^3 + y^3 + 3xy \times 2$, so $x^3 + y^3 + 6xy = 8$ for every such pair. In general:

$$a + b = k \Rightarrow a^3 + b^3 + 3kab = k^3 \qquad a - b = k \Rightarrow a^3 - b^3 - 3kab = k^3$$

[SSC Tip] The middle part of $(a + b)^3$ is $3ab(a + b)$, two pieces multiplied. Writing only $3ab$ is correct only when $a + b = 1$. Quick test of any expansion: put $a = b = 1$. $(1 + 1)^3 = 8$, and $1 + 3 + 3 + 1 = 8$ too.

Cubes from a Sum, a Difference and a Product

Most two-letter cube questions need only these two lines:

$$a^3 + b^3 = (a + b)^3 - 3ab(a + b) \qquad a^3 - b^3 = (a - b)^3 + 3ab(a - b)$$

Straight use.

• $a + b = 8$ and $ab = 7$: $a^3 + b^3 = 512 - 3 \times 7 \times 8 = 512 - 168 = 344$. (7 and 1: $343 + 1 = 344$.)
• $a - b = 4$ and $ab = 5$: $a^3 - b^3 = 64 + 3 \times 5 \times 4 = 64 + 60 = 124$. (5 and 1: $125 - 1 = 124$.)

Notice the sign. For the sum of cubes you take $3ab(a + b)$ away; for the difference of cubes you add $3ab(a - b)$.

When $a^2 + b^2$ is given instead of $ab$. Find $ab$ first, or use the bracket form, which is often quicker. With $a + b = 9$ and $a^2 + b^2 = 45$:

• Route 1: $2ab = 81 - 45 = 36$, so $ab = 18$ and $a^3 + b^3 = 729 - 3 \times 18 \times 9 = 243$.
• Route 2: $a^3 + b^3 = (a + b)(a^2 + b^2 - ab) = 9 \times (45 - 18) = 243$.

Finding the product from the cubes. If $a + b = 5$ and $a^3 + b^3 = 65$, then $125 - 3ab \times 5 = 65$, so $15ab = 60$ and $ab = 4$.

Going backwards to the sum. "If $a^3 + b^3 = 28$ and $ab = 3$, find $a + b$." Call the sum $s$. Then $a^3 + b^3 = s^3 - 3abs$ gives

$$s^3 - 9s = 28$$

Do not try to solve this as a cubic. Put in small whole numbers, or the options: $s = 4$ gives $64 - 36 = 28$. So $a + b = 4$, and then $(a - b)^2 = 16 - 12 = 4$ gives $a - b = \pm 2$. (The numbers are 3 and 1.) The same trick works for the difference: if $a^3 - b^3$ and $ab$ are given, put $d = a - b$ and solve $d^3 + 3abd = a^3 - b^3$ by trying values.

Real numbers again. After finding the sum, check that $s^2 \ge 4ab$, otherwise $a$ and $b$ cannot both be real. Here $16 \ge 12$, so all is well.

Scaled Forms and Rectangle Questions

SSC often hides the identity by putting numbers in front of the letters: "If $2x + 5y = 12$ and $xy = 2$, find $8x^3 + 125y^3$." The trick is to treat $2x$ as one term and $5y$ as the other.

Call them $A = 2x$ and $B = 5y$. Then:

• $A + B = 12$
• $AB = 2x \times 5y = 10xy = 20$
• $8x^3 + 125y^3 = A^3 + B^3$ and $4x^2 + 25y^2 = A^2 + B^2$

Now it is the plain two-letter problem:

$$A^2 + B^2 = 144 - 2 \times 20 = 104 \qquad A^3 + B^3 = 12^3 - 3 \times 20 \times 12 = 1728 - 720 = 1008$$

Check: $x = 5$ and $y = \frac{2}{5}$ fit both conditions, and $1000 + 8 = 1008$.

The identity itself also scales. $(2x + 5y)^3 = 8x^3 + 125y^3 + 30xy(2x + 5y)$, because $3AB = 3 \times 2x \times 5y = 30xy$.

The difference version. "If $2x - y = 5$ and $8x^3 - y^3 = 335$, where $x$ and $y$ are positive, find $4x^2 - y^2$." Take $A = 2x$ and $B = y$:

1. $A^3 - B^3 = (A - B)(A^2 + AB + B^2)$, so $A^2 + AB + B^2 = 335 \div 5 = 67$.
2. $(A - B)^2 = A^2 - 2AB + B^2 = 25$. Take this from step 1: $3AB = 42$, so $AB = 14$.
3. $(A + B)^2 = (A - B)^2 + 4AB = 25 + 56 = 81$, so $A + B = 9$ (positive values).
4. $4x^2 - y^2 = (A - B)(A + B) = 5 \times 9 = 45$.

The word "positive" matters: $A = -2$, $B = -7$ also fits the two conditions and would give $-45$.

Rectangles. For a rectangle with length $l$ and breadth $b$, the perimeter gives $l + b$, the area gives $lb$, and the diagonal is $\sqrt{l^2 + b^2}$. So a rectangle question is really a sum-and-product question.

A rectangle has perimeter 46 cm and area 120 sq cm. Then $l + b = 23$ and $lb = 120$, so $l^2 + b^2 = 529 - 240 = 289$ and the diagonal is $\sqrt{289} = 17$ cm. The sides are 15 cm and 8 cm, and you never had to find them.

[SSC Tip] The product of the scaled terms is the number times $xy$: $2x \times 5y = 10xy$. Using plain $xy$ as the product is the standard slip, and SSC keeps the answer it produces among the options. In rectangle questions, halve the perimeter before you square it.

Factorisation: Common Factors, Grouping and Special Forms

Factorising means writing an expression as a product. It is expanding in reverse, so every identity above becomes a factorising tool. Work through these checks in order.

1. Take out the common factor first. Take the HCF of the numbers and the lowest power of each letter that appears in every term:

$$12x^3y - 18x^2y^2 = 6x^2y(2x - 3y)$$

2. Grouping (four terms). Pair the terms so that each pair has a common factor and the brackets left over match:

$$x^3 - 4x^2 + 3x - 12 = x^2(x - 4) + 3(x - 4) = (x - 4)(x^2 + 3)$$

If the brackets do not match, pair the terms differently. $ab + cd + ac + bd$ does not group as written, but $(ab + ac) + (bd + cd) = a(b + c) + d(b + c) = (a + d)(b + c)$.

3. Difference of squares. $25x^2 - 49y^2 = (5x - 7y)(5x + 7y)$. Often one of the squares is hidden as three terms:

$$4x^2 - 4x + 1 - 9y^2 = (2x - 1)^2 - (3y)^2 = (2x - 1 - 3y)(2x - 1 + 3y)$$

Look for a perfect square of three terms sitting next to a lone square with a minus sign.

4. Perfect squares. If the first and last terms are squares and the middle is twice their product, fold it up: $9x^2 - 30xy + 25y^2 = (3x - 5y)^2$.

5. Sum and difference of cubes.

$$27x^3 + 8 = (3x + 2)(9x^2 - 6x + 4) \qquad x^3 - 64y^3 = (x - 4y)(x^2 + 4xy + 16y^2)$$

The three-term bracket $a^2 \pm ab + b^2$ never factorises further with real numbers, so stop there.

6. [Tier 2] Sixth powers. For $x^6 - y^6$, use squares first and cubes second:

$$x^6 - y^6 = (x^3 - y^3)(x^3 + y^3) = (x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2)$$

Doing cubes first gives $(x^2 - y^2)(x^4 + x^2y^2 + y^4)$, which is correct but not finished, and the $a^4 + a^2b^2 + b^4$ factorisation from the part on fourth powers is needed to finish it.

Before you pick a factorised option, check it with one value. Put $x = 1$ (and $y = 1$) in the question and in the option. $x^3 - 4x^2 + 3x - 12$ at $x = 1$ is $-12$, and $(x - 4)(x^2 + 3)$ at $x = 1$ is $(-3)(4) = -12$. A sign slip in an option shows up straight away.

Splitting the Middle Term

A quadratic like $x^2 + 7x + 12$ often splits into two brackets. Look for two numbers whose product is the last term and whose sum is the middle coefficient. Here 3 and 4 work, so $x^2 + 7x + 12 = (x + 3)(x + 4)$.

Signs. The signs of the last and middle terms tell you the signs of the two numbers.

Last term	Middle term	The two numbers
positive	positive	both positive
positive	negative	both negative
negative	positive	the bigger one positive
negative	negative	the bigger one negative

So $x^2 - 2x - 15$ needs product $-15$ and sum $-2$: the numbers are $-5$ and 3, giving $(x - 5)(x + 3)$.

When $x^2$ has a coefficient. For $ax^2 + bx + c$, multiply $a$ and $c$, find two numbers with that product and sum $b$, split the middle term with them and group.

Take $6x^2 + 11x - 10$. Here $ac = -60$, and 15 and $-4$ have product $-60$ and sum 11:

$$6x^2 + 15x - 4x - 10 = 3x(2x + 5) - 2(2x + 5) = (3x - 2)(2x + 5)$$

Two letters. The method does not change. $2x^2 + 5xy - 12y^2$ has $ac = -24$, and 8 and $-3$ work:

$$2x^2 + 8xy - 3xy - 12y^2 = 2x(x + 4y) - 3y(x + 4y) = (2x - 3y)(x + 4y)$$

A bracket in place of $x$. $(x + 1)^2 - 5(x + 1) + 6$ is a quadratic in $t = x + 1$: $t^2 - 5t + 6 = (t - 2)(t - 3)$. Putting $t$ back gives $(x - 1)(x - 2)$.

A square in place of $x$. $x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4) = (x - 1)(x + 1)(x - 2)(x + 2)$. Split as if $x^2$ were the letter, then use the difference of squares on each bracket.

"Which of these is a factor?" Factorise and look. If the options are brackets like $3x - 4$ and $3x + 4$, the sign is what is being tested, so multiply your two brackets back to check the middle term.

If no pair of whole numbers works, the quadratic does not split nicely; solving it by the formula belongs to the Polynomials and Quadratic Equations section.

When $ac$ is large, list its factor pairs starting from the middle. For $ac = -120$ and middle $-7$, the pair must differ by 7: 8 and 15 do, so the numbers are 8 and $-15$. With a negative last term, the middle coefficient is the difference of the pair, so look for that difference straight away.

Simplifying Algebraic Fractions, and Checking with Simple Values

To simplify a fraction, factorise the top and the bottom completely, then cancel the brackets they share. Never cancel single terms across a plus or minus sign: in $\frac{x + 4}{x + 2}$ nothing cancels.

Difference of squares over one of its brackets.

$$\frac{16x^2 - 9y^2}{4x + 3y} = \frac{(4x - 3y)(4x + 3y)}{4x + 3y} = 4x - 3y$$

A full cube over a difference of squares. $8x^3 - 12x^2 + 6x - 1$ is $(2x - 1)^3$ (from the cube identities above), and $4x^2 - 1 = (2x - 1)(2x + 1)$:

$$\frac{8x^3 - 12x^2 + 6x - 1}{4x^2 - 1} = \frac{(2x - 1)^3}{(2x - 1)(2x + 1)} = \frac{(2x - 1)^2}{2x + 1}$$

Cubes over the matching bracket. $\frac{a^3 - b^3}{a^2 + ab + b^2} = a - b$ and $\frac{a^3 + b^3}{a^2 - ab + b^2} = a + b$. With scaled terms, $\frac{a^3 - 27b^3}{a^2 + 3ab + 9b^2} = a - 3b$, since $27b^3 = (3b)^3$.

Products of fractions. Factorise everything first, then cancel across the whole product:

$$\frac{a^2 - b^2}{a^3 - b^3} \times \frac{a^2 + ab + b^2}{a + b} = \frac{(a - b)(a + b)}{(a - b)(a^2 + ab + b^2)} \times \frac{a^2 + ab + b^2}{a + b} = 1$$

Adding fractions. Use the product of the bottoms when they share nothing: $\frac{1}{x - 2} - \frac{1}{x + 2} = \frac{(x + 2) - (x - 2)}{x^2 - 4} = \frac{4}{x^2 - 4}$.

Checking with simple values. An identity holds for every allowed value, so you can test the question and the options with numbers that make the work tiny. Good choices are $a = 1, b = 0$, or $a = 2, b = 1$.

"$\frac{(a + b)^2 - (a - b)^2}{a^2b + ab^2}$ equals: (a) $\frac{4}{a + b}$ (b) $\frac{2}{a + b}$ (c) $4ab$ (d) $a + b$." Here $b = 0$ would make the bottom zero, so try $a = b = 1$: the question gives $\frac{4 - 0}{2} = 2$. Options (a) and (d) both give 2. Break the tie with $a = 2, b = 1$: the question gives $\frac{9 - 1}{6} = \frac{4}{3}$, which matches (a) but not (d), which gives 3. So the answer is (a), and the algebra agrees: the top is $4ab$ and the bottom is $ab(a + b)$.

Avoid values that make any bottom zero ($x = 2$ in $\frac{4}{x^2 - 4}$) or make two letters equal when the question divides by their difference. If two options survive one test, a second pair of values settles it.

Solved Examples

Question 1

Simplify $(5x - 2y)^2 + 20xy$. (a) $25x^2 - 4y^2$ (b) $25x^2 + 4y^2 + 40xy$ (c) $25x^2 + 4y^2$ (d) $5x^2 + 2y^2$

Answer. Open the square first: $(5x - 2y)^2 = 25x^2 - 2 \times 5x \times 2y + 4y^2 = 25x^2 - 20xy + 4y^2$. Adding $20xy$ wipes out the middle term, so we are left with (c) $25x^2 + 4y^2$.

Question 2

What is the value of $\frac{(4a + 3b)^2 - (4a - 3b)^2}{ab}$? (a) $24$ (b) $48$ (c) $12$ (d) $4$

Answer. This is $(A + B)^2 - (A - B)^2 = 4AB$ with $A = 4a$ and $B = 3b$. So the top is $4 \times 4a \times 3b = 48ab$, and dividing by $ab$ gives (b) $48$. Using $2AB$ by mistake gives 24.

Question 3

If $(x + 7)(x - 3) = x^2 + px + q$, what is $p + q$? (a) $17$ (b) $-25$ (c) $25$ (d) $-17$

Answer. For $(x + a)(x + b)$ the middle coefficient is $a + b$ and the last term is $ab$. Here the two numbers are 7 and $-3$, so $p = 7 - 3 = 4$ and $q = 7 \times (-3) = -21$. Then $p + q = 4 - 21$, which is (d) $-17$.

Question 4

For what positive value of $k$ is $9x^2 - kx + 49$ a perfect square? (a) 42 (b) 21 (c) 84 (d) 14

Answer. $9x^2 = (3x)^2$ and $49 = 7^2$, so the square has to be $(3x - 7)^2 = 9x^2 - 42x + 49$. The middle term is $2 \times 3x \times 7 = 42x$, so $k$ is (a) 42.

Question 5

If $a + b = 16$ and $ab = 55$, what is $a^2 + b^2$? (a) 201 (b) 146 (c) 366 (d) 36

Answer. $a^2 + b^2 = (a + b)^2 - 2ab = 256 - 110$, which is (b) 146. (11 and 5 fit: $121 + 25 = 146$.) The option 36 is $(a - b)^2$, not what was asked.

Question 6

If $a - b = 4$ and $ab = 21$, where $a$ and $b$ are positive, what is $a + b$? (a) $\sqrt{58}$ (b) $\sqrt{37}$ (c) $10$ (d) $100$

Answer. Use $(a + b)^2 = (a - b)^2 + 4ab = 16 + 84 = 100$. Both numbers are positive, so $a + b$ is positive too: (c) $10$. Quick check: 7 and 3 differ by 4 and multiply to 21.

Question 7

If $a + b = 9$ and $a^2 + b^2 = 53$, what is $\frac{a}{b} + \frac{b}{a}$? (a) $\frac{14}{53}$ (b) $\frac{81}{14}$ (c) $\frac{53}{28}$ (d) $\frac{53}{14}$

Answer. First get $ab$: $2ab = (a + b)^2 - (a^2 + b^2) = 81 - 53 = 28$, so $ab = 14$. Then $\frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab}$, which is (d) $\frac{53}{14}$.

Question 8

If $a^2 + b^2 = 130$ and $ab = 63$, where $a > b > 0$, what is $a^2 - b^2$? (a) 32 (b) 16 (c) 67 (d) 2

Answer. $a^2 - b^2 = (a + b)(a - b)$, so we need both brackets. $(a + b)^2 = 130 + 126 = 256$ gives $a + b = 16$, and $(a - b)^2 = 130 - 126 = 4$ gives $a - b = 2$ (positive, because $a > b$). Multiply them: (a) 32. The numbers are 9 and 7, and $81 - 49 = 32$ agrees.

Question 9

A rectangle has a perimeter of 82 cm and an area of 420 sq cm. What is the length of its diagonal? (a) 41 cm (b) 29 cm (c) 21 cm (d) 1 cm

Answer. Half the perimeter is the sum of the sides, so $l + b = 41$, and the area gives $lb = 420$. The diagonal is $\sqrt{l^2 + b^2}$, and $l^2 + b^2 = (l + b)^2 - 2lb = 1681 - 840 = 841$. So the diagonal is $\sqrt{841}$, which is (b) 29 cm. (The sides are 21 cm and 20 cm.) Taking away $4lb$ instead of $2lb$ gives 1 cm, which is $l - b$.

Question 10

If $a + b = 5$ and $ab = 3$, what is $a^4 + b^4$? (a) 361 (b) 355 (c) 343 (d) 352

Answer. Do it in two steps. $a^2 + b^2 = 25 - 6 = 19$. Then $a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 = 361 - 2 \times 9$, which is (c) 343. The easy slip is taking away $2ab$ instead of $2a^2b^2$, which gives 355.

Question 11

If $a^4 + a^2b^2 + b^4 = 273$ and $a^2 + ab + b^2 = 21$, what is $a^2 + b^2$? (a) 13 (b) 4 (c) 34 (d) 17

Answer. $a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2)$, so $a^2 - ab + b^2 = 273 \div 21 = 13$. Adding the two brackets gives $2(a^2 + b^2) = 21 + 13 = 34$, so $a^2 + b^2$ is (d) 17. (Taking one bracket from the other gives $2ab = 8$, so $ab = 4$.)

Question 12

If $a + b = 5$ and $a - b = 1$, what is $a^4 - b^4$? (a) 65 (b) 5 (c) 13 (d) 130

Answer. $a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)$. The second bracket is $(a + b)(a - b) = 5 \times 1 = 5$. For the first, $(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$, so $a^2 + b^2 = \frac{25 + 1}{2} = 13$. So $a^4 - b^4 = 13 \times 5$, which is (a) 65. (Here $a = 3$ and $b = 2$, and $81 - 16 = 65$.)