Algebraic identities in numerical simplification — notes, question types & practice · Simplification

01

Overview

A cube of $9.3$ looks like hard work until you notice the shape underneath. $\dfrac{9.3^3 - 3.3^3}{9.3^2 + 9.3 \times 3.3 + 3.3^2}$ collapses to $9.3 - 3.3 = 6$ because the denominator is exactly the second factor of $a^3 - b^3$ . That is the whole game: spot the identity, name $a$ and $b$ , substitute at the end.

02

The identities that do the work

Identity	Reads as
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$	sum of cubes
$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$	difference of cubes
$a^2 - b^2 = (a+b)(a-b)$	difference of squares
$(a+b)^2 + (a-b)^2 = 2(a^2+b^2)$	squares combine
$(a+b)^2 - (a-b)^2 = 4ab$	squares subtract

Rule: In an identity fraction, the sign in the denominator is opposite to the sign on top. Minus below pairs with plus above.

03

Identity fractions

Read the denominator first, because it tells you what cancels. If the top is a difference of cubes and the bottom has a plus middle term, the whole fraction is $a - b$ . Check $\dfrac{12.4^3 - 2.4^3}{12.4^2 + 12.4 \times 2.4 + 2.4^2}$ : the bottom is the plus version, so the value is $12.4 - 2.4 = 10$ . Never cube the decimals.

04

Squares that sit near each other

$115^2 - 85^2 = (115 - 85)(115 + 85) = 30 \times 200 = 6000$ . Whenever two squares sit close together, factor instead of squaring. The same identity multiplies numbers around a round value: $997 \times 1003 = 1000^2 - 3^2 = 999991$ .

Tip: Look for a shared centre. Write the pair as centre minus gap and centre plus gap, then use centre squared minus gap squared.

05

The x plus one over x ladder

If $x + \dfrac{1}{x} = k$ , then $x^2 + \dfrac{1}{x^2} = k^2 - 2$ , and $x^3 + \dfrac{1}{x^3} = k^3 - 3k$ . With the minus version the twos flip: if $x - \dfrac{1}{x} = k$ , then $x^2 + \dfrac{1}{x^2} = k^2 + 2$ . So $x + \dfrac{1}{x} = 4$ gives $16 - 2 = 14$ for the square, and $x + \dfrac{1}{x} = 6$ gives $36 - 2 = 34$ .

06

Cubes from a sum and a product

Given $a+b$ and $ab$ , never hunt for $a$ and $b$ . Use $a^2 + b^2 = (a+b)^2 - 2ab$ and $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$ . With $a+b = 7$ and $ab = 12$ : $a^3 + b^3 = 343 - 252 = 91$ , which is $3^3 + 4^3$ .

07

Zero-sum cubes

If $a + b + c = 0$ , then $a^3 + b^3 + c^3 = 3abc$ . Test the bases before cubing anything: $17 - 10 - 7 = 0$ , so $17^3 + (-10)^3 + (-7)^3 = 3 \times 17 \times 10 \times 7 = 3570$ . Two negative bases multiply to a positive here, so the sign needs one careful look.

Watch: Add the three bases first. If they do not sum to zero, this shortcut does not apply and the numbers must be cubed honestly.

08

A safe attacking order

Read the denominator of any fraction; it names the identity.
Match the middle sign to choose the plus or minus version.
Keep $a$ and $b$ symbolic until the expression is factored.
Substitute the numbers in the last line only.

09

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Identity fraction: cubes over the quadratic factor

How to spot it:

A fraction with cubed decimals on top and three terms with a middle product below.

\dfrac{a^3 \pm b^3}{a^2 \mp ab + b^2} = a \pm b

Method

Read the sign of the middle term in the denominator.
Pick the matching identity: minus below pairs with plus above.
Cancel the quadratic factor.
Add or subtract the two bases; that is the answer.

Why it works:

The denominator is literally the second factor of the numerator's factorisation.

Try this

Find the value of $\dfrac{9.3^3 - 3.3^3}{9.3^2 + 9.3 \times 3.3 + 3.3^2}$ .

Show solution

Bottom is the plus version, so the fraction equals $a - b$ .
$a - b = 9.3 - 3.3$ .
$= 6$ .

Answer

6

Type 2very common2 practice Q

Squares near each other; round-about products

How to spot it:

Two squares of close numbers, or a product of two numbers placed symmetrically around a round value.

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

Method

Find the centre: the average of the two numbers.
Find the gap: the distance from the centre to either number.
Multiply the sum by the difference, or use centre squared minus gap squared.

Why it works:

Factoring avoids squaring three-digit numbers by hand.

Try this

Find $115^2 - 85^2$ .

Show solution

$115 - 85 = 30$ and $115 + 85 = 200$ .
$30 \times 200 = 6000$ .

Answer

6000

Type 3very common2 practice Q

The x plus one over x ladder

How to spot it:

A given value of $x + \frac{1}{x}$ or $x - \frac{1}{x}$ , and a target square, cube or fourth power.

x + \tfrac{1}{x} = k \Rightarrow x^2 + \tfrac{1}{x^2} = k^2 - 2

Method

Square the given relation; the cross term is $\pm 2$ .
Plus version subtracts 2, minus version adds 2.
For cubes use $k^3 \mp 3k$ ; for fourth powers, square the square result and subtract 2.

Why it works:

Squaring $x \pm \frac{1}{x}$ produces the cross term $2$ or $-2$ beside the wanted square-sum.

Try this

If $x + \dfrac{1}{x} = 6$ , find $x^2 + \dfrac{1}{x^2}$ .

Show solution

Square: $6^2 = 36$ .
Subtract the cross term: $36 - 2 = 34$ .

Answer

34

Type 4very common2 practice Q

Given a + b and ab

How to spot it:

The pair sum and product are given; the target is a square-sum, cube-sum or a similar symmetric value.

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

Method

Write the target using only $(a+b)$ and $ab$ .
Substitute the two given numbers.
If the numbers themselves are wanted, spot the small integer pair.

Why it works:

Every symmetric expression in two letters is a function of their sum and product.

Try this

If $a + b = 8$ and $ab = 15$ , find $a^3 + b^3$ .

Show solution

$(a+b)^3 = 512$ .
$3ab(a+b) = 3 \times 15 \times 8 = 360$ .
$512 - 360 = 152$ , which is $3^3 + 5^3$ .

Answer

152

Type 5common2 practice Q

Zero-sum cubes

How to spot it:

Three cubes, often with negative or fractional bases, whose bases add to zero.

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

Method

Add the three bases and confirm the sum is zero.
Multiply the three bases together.
Triple the product, watching the signs.

Why it works:

The identity for $a^3+b^3+c^3 - 3abc$ carries the factor $(a+b+c)$ , which is zero here.

Try this

Find $(-13)^3 + 9^3 + 4^3$ .

Show solution

$-13 + 9 + 4 = 0$ , so the rule applies.
$(-13) \times 9 \times 4 = -468$ .
$3 \times (-468) = -1404$ .

Answer

-1404

10

Formula sheet

Sum of cubes

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

Difference of cubes

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

Difference of squares

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

Cube of sum

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

Squares combine

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

Squares subtract

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

Zero-sum cubes

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

Square-sum from pair facts

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

11

Shortcuts that save time

⚡ Pattern-match the fraction

Cubes on top and three terms with a middle product below mean a sum or difference of cubes. The fraction is just a plus or minus of the bases.

Example

Simplify (0.7 x 0.7 x 0.7 - 0.3 x 0.3 x 0.3) over (0.7 x 0.7 + 0.7 x 0.3 + 0.3 x 0.3).

Show solution

Shape: $\dfrac{a^3 - b^3}{a^2 + ab + b^2}$ with $a = 0.7$ , $b = 0.3$ .
Value $= a - b = 0.7 - 0.3$ .
$0.7 - 0.3 = 0.4 = \dfrac{2}{5}$ .

Answer

2/5

⚡ Zero-sum check before cubing

Add the three bases. If they sum to zero, the cubes sum to three times their product, signs included.

Example

Find (-16)^3 + 9^3 + 7^3.

Show solution

$-16 + 9 + 7 = 0$ , so the zero-sum rule applies.
Value $= 3 \times (-16) \times 9 \times 7$ .
$3 \times (-1008) = -3024$ .

Answer

-3024

⚡ Products around a round number

Write the pair as centre minus gap and centre plus gap. The product is centre squared minus gap squared.

Example

Find 997 x 1003.

Show solution

Centre $1000$ , gap $3$ .
$1000^2 - 3^2 = 1000000 - 9$ .
$= 999991$ .

Answer

999991

12

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Pairing a sum-of-cubes top with a plus middle term below.

The middle sign below is opposite to the sign on top.

Mistake 02

Cubing the decimals in an identity fraction.

Name a and b, cancel the factor, subtract or add the bases.

Mistake 03

Writing $(a+b)^3 = a^3 + b^3$ .

The cube of a sum carries the extra term $3ab(a+b)$ .

Mistake 04

Using $k^2 + 2$ when $x + 1/x = k$ is given.

Plus version subtracts 2; minus version adds 2.

Mistake 05

Applying the zero-sum rule without adding the bases.

Check $a + b + c = 0$ first; otherwise cube honestly.

13

Quick revision

Read this the night before the exam.

$\dfrac{a^3 \pm b^3}{a^2 \mp ab + b^2} = a \pm b$ .
$a^2 - b^2 = (a+b)(a-b)$ for near squares and round-about products.
$x + \dfrac{1}{x} = k$ : square gives $k^2 - 2$ , cube gives $k^3 - 3k$ .
$a^3 + b^3 = (a+b)^3 - 3ab(a+b)$ from a sum and a product.
$a + b + c = 0$ gives $a^3 + b^3 + c^3 = 3abc$ .
Substitute numbers only in the last line.

14

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.