Basic algebraic identities — notes, question types & practice · Algebra

01

Overview

Five identities answer almost everything here:

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

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

Read each one both ways. Expanding is one direction; spotting the factor inside a giant expression is the other, and that is the direction the exam pays for.

02

Build upward from a sum and a product

When the question gives $a+b$ and $ab$ , never solve for $a$ and $b$ . Climb instead:

a^2 + b^2 = (a+b)^2 - 2ab \qquad a^3 + b^3 = (a+b)^3 - 3ab(a+b) \qquad a^4 + b^4 = (a^2+b^2)^2 - 2a^2b^2

With $a+b = 5$ and $ab = 6$ : first $a^2 + b^2 = 25 - 12 = 13$ , then $a^3 + b^3 = 125 - 90 = 35$ , then $a^4 + b^4 = 169 - 72 = 97$ . Each rung reuses the one below it.

Rule: Every symmetric expression in two letters is a function of $a+b$ and $ab$ alone. Two given numbers are always enough.

03

Big numbers hide an identity

A fraction with cubed terms is usually an identity fraction. Read the denominator first: a minus middle term pairs with a sum of cubes, a plus middle term with a difference of cubes. So $\dfrac{94^3 + 6^3}{94^2 - 94 \times 6 + 6^2}$ is just $94 + 6 = 100$ . Squares stack the same way: $873^2 + 127^2 + 873 \times 127$ is $(a+b)^2 - ab$ with $a + b = 1000$ , giving $1000000 - 110871 = 889129$ .

04

Squares that must be zero

An equation like $x^2 + y^2 - 12x + 4y + 40 = 0$ is two perfect squares in disguise. Group the $x$ terms and the $y$ terms: $x^2 - 12x + 36$ and $y^2 + 4y + 4$ , so $(x-6)^2 + (y+2)^2 = 0$ . A square is never negative, so both are zero: $x = 6$ , $y = -2$ .

Watch: Complete both squares and check that the constants exactly absorb the loose number. If they do not, the equation has other solutions and this route is wrong.

05

Products around a round base

Write each number as base plus or minus a small offset. $205^2 - 195^2 = (205 - 195)(205 + 195) = 10 \times 400 = 4000$ . And $103 \times 97 = (100+3)(100-3) = 10000 - 9 = 9991$ . No long multiplication anywhere.

06

Choose expressions by value-putting

When the options are themselves expressions, substitute small numbers into the question and into every option. Only the right option survives. For $(a+b)^2 - (a-b)^2$ with $a = 1$ , $b = 2$ : the value is $9 - 1 = 8$ , and $4ab = 8$ matches.

Tip: Use two different non-zero numbers, and test a second pair if two options tie. Avoid values that make a denominator zero.

07

An order that works

Read the denominator of any fraction; it names the identity.
Match the middle sign before choosing plus or minus forms.
Keep the letters symbolic until the last line.
Substitute once, at the end.

08

Question types you will see

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

Type 1very common4 practice Q

Symmetric expressions from a sum and a product

How to spot it:

Two facts like $a+b=9$ and $ab=20$ are given, and a higher symmetric expression is asked.

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

Method

Never solve for the two letters.
Write the target using only $(a+b)$ and $ab$ .
Substitute the two given numbers and finish.

Why it works:

Every symmetric polynomial in two letters is built from their sum and product.

Try this

If $a + b = 9$ and $ab = 20$ , find $a^3 + b^3$ .

Show solution

$(a+b)^3 = 729$ .
$3ab(a+b) = 3 \times 20 \times 9 = 540$ .
$729 - 540 = 189$ (the numbers are $4$ and $5$ ).

Answer

189

Type 2very common2 practice Q

Big-number simplification by identity

How to spot it:

Fractions with cubed large numbers over a trinomial, or stacks of squares like $a^2+b^2+ab$ .

\dfrac{p^3 \pm q^3}{p^2 \mp pq + q^2} = p \pm q

Method

Read the denominator's middle sign.
Pick the cube identity with the opposite sign.
The fraction collapses to the sum or difference of the bases.

Why it works:

The denominator is exactly the second factor of the cube identity.

Try this

Find the value of $\dfrac{94^3 + 6^3}{94^2 - 94 \times 6 + 6^2}$ .

Show solution

Bottom has the minus middle term, so it pairs with a sum of cubes.
Value $= 94 + 6$ .
$= 100$ .

Answer

100

Type 3very common2 practice Q

Sum of squares equal to zero

How to spot it:

One equation in two or three variables, with squared and linear terms and a loose constant.

(x-a)^2 + (y-b)^2 = 0 \Rightarrow x = a,\ y = b

Method

Group the $x$ terms and the $y$ terms.
Complete each square; the constants must absorb the loose number.
Each square is zero, so read off both variables.
Compute the asked combination.

Why it works:

A square cannot be negative, so a zero sum of squares pins every letter.

Try this

If $x^2 + y^2 - 12x + 4y + 40 = 0$ , find $x - y$ .

Show solution

$x^2 - 12x + 36$ and $y^2 + 4y + 4$ use up the $40$ exactly.
$(x-6)^2 + (y+2)^2 = 0$ , so $x = 6$ , $y = -2$ .
$x - y = 6 - (-2) = 8$ .

Answer

8

Type 4common2 practice Q

Direct evaluation near a round base

How to spot it:

Products or differences of squares of numbers sitting symmetrically around a round base.

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

Method

Write each number as base plus or minus an offset.
Apply the difference of squares.
Compute in the base, which squares easily.

Why it works:

The round base squares in one step, and the small offsets square to nothing.

Try this

Find the value of $205^2 - 195^2$ .

Show solution

$205 - 195 = 10$ and $205 + 195 = 400$ .
$10 \times 400 = 4000$ .

Answer

4000

Type 5common

Pick the matching expression by value-putting

How to spot it:

The options are algebraic expressions, not numbers, and the question asks which one is equal.

\text{substitute small numbers in question and options}

Method

Pick two different small non-zero numbers.
Evaluate the question expression.
Evaluate every option on the same numbers.
If two options survive, test a second pair.

Why it works:

An identity holds for every value, so one clean substitution filters all wrong options.

Try this

$(a+b)^2 - (a-b)^2$ is equal to which of $2ab$ , $4ab$ , $2(a^2+b^2)$ , $a^2-b^2$ ?

Show solution

Put $a = 1$ , $b = 2$ : value $= 9 - 1 = 8$ .
Options give $4$ , $8$ , $10$ , $-3$ .
Only $4ab$ gives $8$ ; answer $4ab$ .

Answer

4ab

09

Formula sheet

Square of sum or difference

(a \pm b)^2 = a^2 \pm 2ab + b^2

Difference of squares

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

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)

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

Square of a trinomial

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

Fourth powers from the ladder

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

10

Shortcuts that save time

⚡ Build higher powers from a sum and a product

Chain the rungs: square-sum first, then cube-sum, then fourth powers. The individual letters are never needed.

Example

If a + b = 5 and ab = 6, find a^3 + b^3.

Show solution

$(a+b)^3 = 125$ .
$3ab(a+b) = 3 \times 6 \times 5 = 90$ .
$a^3 + b^3 = 125 - 90 = 35$ (the numbers are $2$ and $3$ ).

Answer

35

⚡ Recognise the identity inside decimals

Cubed decimals over a trinomial mean a cube identity. The value is the sum or difference of the bases.

Example

Simplify (4.7^3 + 2.3^3) / (4.7^2 - 4.7 x 2.3 + 2.3^2).

Show solution

Bottom has the minus middle term, so it pairs with a sum of cubes.
Value $= 4.7 + 2.3$ .
$= 7$ .

Answer

7

⚡ Value-putting among expression options

Put small numbers into the question and each option. Wrong options die in one round; ties die in a second.

Example

(a + b)^2 - (a - b)^2 equals which of: 2ab, 4ab, 2(a^2+b^2), a^2-b^2?

Show solution

Put $a = 1$ , $b = 2$ : value $= 9 - 1 = 8$ .
Options give $4$ , $8$ , $10$ , $-3$ .
Only $4ab$ matches, so it is the answer.

Answer

4ab

11

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Writing $(a+b)^2 = a^2 + b^2$ .

The cross term $2ab$ is the most commonly lost mark; check with $a = b = 1$ .

Mistake 02

Pairing a sum of cubes with the plus middle trinomial.

The middle sign in the factor is opposite to the sign between the cubes.

Mistake 03

Solving for a and b individually when a+b and ab are given.

Build the target from the sum and the product; the letters never surface.

Mistake 04

Value-putting with a = b or with zero.

Use two different non-zero values so options cannot collapse together.

Mistake 05

Declaring a square-sum equation solved without completing both squares.

The constants must absorb the loose number exactly; then each square is zero.

12

Quick revision

Read this the night before the exam.

$a^2+b^2 = (a+b)^2 - 2ab$ ; $a^3+b^3 = (a+b)^3 - 3ab(a+b)$ .
$\dfrac{a^3 \pm b^3}{a^2 \mp ab + b^2} = a \pm b$ .
$(a+b)^2 - (a-b)^2 = 4ab$ .
Squares summing to zero force each bracket to zero.
Near a round base, split as base plus and minus the offset.
Test expression options by value-putting with two small pairs.

13

Practice: 13 questions

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

Topic test · 13 questions

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