Algebraic Identities, Expansions and Factorisation | NIMCET

Quick answer: The identities used most in NIMCET are $(a \pm b)^2 = a^2 \pm 2ab + b^2$ , $a^2 - b^2 = (a - b)(a + b)$ , $(a \pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3$ , $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$ and $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ . The remainder theorem says $p(x)$ divided by $(x - a)$ leaves remainder $p(a)$ ; if $p(a) = 0$ , then $(x - a)$ is a factor.

Core identities

Identity	Expansion
$(a + b)^2$	$a^2 + 2ab + b^2$
$(a - b)^2$	$a^2 - 2ab + b^2$
$a^2 - b^2$	$(a - b)(a + b)$
$(a + b + c)^2$	$a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$(a + b)^3$	$a^3 + b^3 + 3ab(a + b)$
$(a - b)^3$	$a^3 - b^3 - 3ab(a - b)$
$a^3 + b^3$	$(a + b)(a^2 - ab + b^2)$
$a^3 - b^3$	$(a - b)(a^2 + ab + b^2)$
$a^4 + a^2b^2 + b^4$	$(a^2 + ab + b^2)(a^2 - ab + b^2)$

The three-variable cube identity

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

Two consequences appear often:

If $a + b + c = 0$ , then $a^3 + b^3 + c^3 = 3abc$ .
$a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right] \ge 0$ .

Useful derived results

Given	Result
$x + \frac{1}{x} = k$	$x^2 + \frac{1}{x^2} = k^2 - 2$
$x + \frac{1}{x} = k$	$x^3 + \frac{1}{x^3} = k^3 - 3k$
$x - \frac{1}{x} = k$	$x^2 + \frac{1}{x^2} = k^2 + 2$
$a + b$ and $ab$ known	$a^2 + b^2 = (a + b)^2 - 2ab$
$a + b$ and $ab$ known	$(a - b)^2 = (a + b)^2 - 4ab$

Factorisation methods

Method	Example
Common factor	$6x^2 + 9x = 3x(2x + 3)$
Grouping	$x^3 + x^2 + x + 1 = (x + 1)(x^2 + 1)$
Splitting the middle term	$x^2 + 5x + 6 = (x + 2)(x + 3)$
Difference of squares	$x^4 - 16 = (x^2 + 4)(x + 2)(x - 2)$
Sum or difference of cubes	$8x^3 - 27 = (2x - 3)(4x^2 + 6x + 9)$
Factor theorem	Find a root, then divide

Remainder and factor theorems

Remainder theorem: the remainder when $p(x)$ is divided by $(x - a)$ is $p(a)$ . For $(ax - b)$ , it is $p\left(\frac{b}{a}\right)$ .
Factor theorem: $(x - a)$ is a factor of $p(x)$ if and only if $p(a) = 0$ .

A quick check: if the coefficients of $p(x)$ add to 0, then $x = 1$ is a root and $(x - 1)$ is a factor.

Worked examples

Example 1: using x + 1/x

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

$3^3 - 3 \cdot 3 = 27 - 9 = 18$ .

Example 2: the a + b + c = 0 shortcut

Evaluate $(-12)^3 + 7^3 + 5^3$ .

Since $-12 + 7 + 5 = 0$ , the value is $3(-12)(7)(5) = -1260$ .

Example 3: remainder

Find the remainder when $x^3 - 2x^2 + 4x - 5$ is divided by $x - 2$ .

$p(2) = 8 - 8 + 8 - 5 = 3$ .

Example 4: factorise a cubic

Factorise $x^3 - 6x^2 + 11x - 6$ .

Step 1: The coefficients add to 0, so $(x - 1)$ is a factor.

Step 2: Dividing gives $x^2 - 5x + 6 = (x - 2)(x - 3)$ .

Answer: $(x - 1)(x - 2)(x - 3)$ .

Summary table

Idea	Key result
$a^3 + b^3$	$(a + b)(a^2 - ab + b^2)$
$a + b + c = 0$	$a^3 + b^3 + c^3 = 3abc$
$x + \frac{1}{x} = k$	$x^3 + \frac{1}{x^3} = k^3 - 3k$
Remainder on dividing by $x - a$	$p(a)$
Sum of coefficients 0	$(x - 1)$ is a factor

Exam tips

Check for $a + b + c = 0$ before computing any sum of three cubes.
In "find the value" problems, express the target in terms of $a + b$ and $ab$ .
Use the remainder theorem instead of long division whenever only the remainder is asked.

These skills support quadratic equations and the rest of the NIMCET Algebra chapter.

Common mistakes

Writing $(a + b)^2 = a^2 + b^2$ .
Getting the middle sign wrong in $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ .
Using $p(-a)$ instead of $p(a)$ for division by $(x - a)$ .
Stopping factorisation too early, as in leaving $x^4 - 16$ as $(x^2 - 4)(x^2 + 4)$ .

Frequently Asked Questions

What is the formula for a³ + b³ + c³ − 3abc?

It equals (a + b + c)(a² + b² + c² − ab − bc − ca).

If a + b + c = 0, what is a³ + b³ + c³?

It equals 3abc.

What is the remainder theorem?

When a polynomial p(x) is divided by (x − a), the remainder is p(a).

How do you find x³ + 1/x³ from x + 1/x?

If x + 1/x = k, then x³ + 1/x³ = k³ − 3k.

What is the factor theorem?

(x − a) is a factor of p(x) exactly when p(a) = 0.