Algebraic Identities

A polynomial identity is an equation that is true for every value of the variables. Identities are not solved , they are used. Each identity is a shortcut: instead of multiplying step by step, you write down the result by reading off the pattern. The identities below are the eight you must know cold.

Definitions

An algebraic identity is an equality between two algebraic expressions that holds for all real values of the variables involved. The basic distributive law $a(b + c) = ab + ac$ is the simplest example; the identities of this lesson are richer consequences.

The eight standard identities

We list each, give a one-line proof or motivation, and a quick example.

Identity 1. $(a + b)^2 = a^2 + 2ab + b^2$.

Proof. Expand: $(a+b)(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2$.

Example. $(x + 3)^2 = x^2 + 6x + 9$.

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

Proof. Replace $b$ by $-b$ in Identity 1.

Example. $(x - 5)^2 = x^2 - 10x + 25$.

Identity 3. $a^2 - b^2 = (a + b)(a - b)$.

Proof. Expand the right side: $a^2 - ab + ab - b^2 = a^2 - b^2$.

Example. $25 - x^2 = (5 - x)(5 + x)$. Also $49^2 - 1^2 = (49+1)(49-1) = 50 \cdot 48 = 2400$.

Identity 4 (linear product). $(x + a)(x + b) = x^2 + (a + b)x + ab$.

Example. $(x + 4)(x - 7) = x^2 + (4 + (-7))x + (4)(-7) = x^2 - 3x - 28$.

Identity 5 (trinomial square). $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$.

Example. $(x + 2 + y)^2 = x^2 + 4 + y^2 + 2(2x + 2y + xy) = x^2 + y^2 + 4 + 4x + 4y + 2xy$.

Identity 6 (cube of a sum/difference). $(a + b)^3 = a^3 + 3a^2 b + 3 a b^2 + b^3 = a^3 + b^3 + 3ab(a + b)$ and $(a - b)^3 = a^3 - 3a^2 b + 3 a b^2 - b^3 = a^3 - b^3 - 3ab(a-b)$.

Example. $(x + 2)^3 = x^3 + 6x^2 + 12x + 8$.

Identity 7 (sum and difference of cubes). $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

Example. $x^3 - 27 = (x - 3)(x^2 + 3x + 9)$.

Identity 8 (three-cube identity). $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.

This stunning factorisation has a useful corollary: if $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Example. If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$. Take $a = 1, b = 2, c = -3$: $1 + 8 - 27 = -18 = 3 \cdot 1 \cdot 2 \cdot (-3)$. Checks out.

Three useful auxiliary observations

• $a^2 + b^2 = (a + b)^2 - 2ab = (a - b)^2 + 2ab$. Very handy when you know $a + b$ and $ab$.
• $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$, an alternative form.
• $a^2 + b^2 + c^2 - ab - bc - ca = \tfrac{1}{2}\!\left[(a-b)^2 + (b-c)^2 + (c-a)^2\right] \ge 0$ for all real $a, b, c$.

Worked examples

Example 1. Expand $(2x + 3y)^2$.

By Identity 1: $(2x)^2 + 2(2x)(3y) + (3y)^2 = 4x^2 + 12xy + 9y^2$.

Example 2. Evaluate $98 \times 102$ without a calculator.

Use Identity 3 backwards: $98 \times 102 = (100 - 2)(100 + 2) = 100^2 - 2^2 = 10000 - 4 = 9996$.

Example 3. Factorise $x^3 + 8$.

By Identity 7: $x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)$.

Example 4. Expand $(x + 2y + 3z)^2$.

By Identity 5: $x^2 + 4y^2 + 9z^2 + 2(x \cdot 2y + 2y \cdot 3z + 3z \cdot x) = x^2 + 4y^2 + 9z^2 + 4xy + 12yz + 6xz$.

Example 5. Without expanding, find the value of $a^3 + b^3 + c^3 - 3abc$ when $a = 5, b = -4, c = -1$.

$a + b + c = 0$, so by Identity 8 the expression equals $0$. (Equivalently, $a^3 + b^3 + c^3 = 3abc = 3 \cdot 5 \cdot (-4) \cdot (-1) = 60$, and $5^3 + (-4)^3 + (-1)^3 = 125 - 64 - 1 = 60$. Checks out.)

Try it yourself

1. Expand $(3x - 4)^2$.
2. Factorise $x^2 - 144$.
3. Expand $(x + 1)^3$.
4. Factorise $27x^3 - 1$.
5. Compute $103 \times 97$ using a difference of squares.
6. Expand $(a + b - c)^2$.
7. Factorise $x^3 + 27y^3$.
8. If $a + b = 7$ and $ab = 12$, find $a^2 + b^2$ and $a^3 + b^3$.
9. Prove that if $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.
10. Compute $25^3 - 75^3 + 50^3$ using Identity 8.

Pitfalls / Insight

• Sign errors are the most common mistake. Especially with $(a - b)^2$ and $(a - b)^3$ , write out the middle terms carefully.
• Don't confuse $a^2 - b^2$ (factorisable) with $a^2 + b^2$ (not factorisable over the reals). Only the difference factors as $(a+b)(a-b)$; the sum cannot be split this way.
• Trinomial identities reward symmetry. Whenever the problem has three variables in a symmetric pattern, look at Identity 5 or 8.

Insight. Every identity here can be derived from the distributive law in a few lines. Memorising them is just a shortcut. The deeper value is in spotting when to use them: the moment you see $(a + b)$ next to $(a - b)$, write down their product as $a^2 - b^2$; the moment you see two cubes added, factor as $(a + b)(a^2 - ab + b^2)$. Pattern recognition is what makes algebra fast.