In this article, we will look at standard identities as well as some other Algebraic identities. In addition, we will attempt to derive these identities without using the binomial theorem. We’ll also look at some solved examples of problems that use these mathematical identities to solve them.

## What is an Algebraic Identity

An **Algebraic identity** is equality, which is true for all values of the variables in the equality. While an equation is true only for certain values of its variables. An equation is not an identity. These identities are used during the factorization of polynomials.

**Why an equation is not an identity?**

x+ 2 =5

Now, this is true for x=3 only. So this is not an identity

$(x+1)^2 = x^2 + 2x +1$

Now this is true for x=0,1,2 –. So this is an identity

The Binomial Theorem is used to derive all of the standard Algebraic Identities and is given as:

\begin{array}{l} (a+b)^{n} =\; ^{n}C_{0}.a^{n}.b^{0} +^{n} C_{1} . a^{n-1} . b^{1} + …….. + ^{n}C_{n-1}.a^{1}.b^{n-1} + ^{n}C_{n}.a^{0}.b^{n}\end{array}

## Standard Identities

There are four standard Identities

**Identity (I)**

$(a + b)^2 = a^2 + 2ab + b^2$**Derivation:**

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

$= a(a + b) + b (a + b)$

$= a^2 + ab + ba + b^2$

$= a^2 + 2ab + b^2$

**Identity (II)**

$(a – b)^2 = a^2 – 2ab + b^2$**Derivation:**

$(a – b)^2 = (a – b) (a – b)$

$= a(a – b) – b (a – b)$

$= a^2 -ab – ba + b^2$

$= a^2 – 2ab + b^2$

**Identity(III)**

$(a + b) (a – b) = a^2– b^2$**Derivation:**

$(a+b)(a – b) = a(a – b) + b (a – b)$

$= a^2 -ab + ba – b^2$

$= a^2 – b^2$

**Identity(IV)**

$(x + a) (x + b) = x^2 + (a + b) x + ab $**Derivation:**

$(x + a) (x + b) = x(x+b) + a (x+b)$

$=x^2 + xb + ax + ab$

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

### Special Cases

**(i) for b =a**

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

$(x+a)^2 = x^2 + 2ax + a^2$

which is the same as identity (I)

** (ii) for b =-c,a=-c**

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

$(x-c)^2 = x^2 – 2cx + c^2$

which is the same as identity (II)

**(iii) for b =-a **

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

$(x+a)(x-a) = x^2 – a^2$

which is the same as identity (III)

**(iv)For b=-b**

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

** (v) For a=-a**

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

**(vi) For a=-a and b=-b**

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

## Other Algebraic Identities

**Identity (V)**

$(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$**Derivation:**

$(x + y + z)^2 =(x+y+z)(x+y+z)$

$=x(x+y+z) + y ( x+y+z ) + z( x+y+z )$

$=x^2 + xy + xz + yx + y^2 + yz + zx + zy +z^2$

$= x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

The RHS side is called the expanded form

**Identity (VI)**

$(x + y)^3 = x^3 + y^3 + 3xy (x + y)$**Derivation:**

$(x + y)^3 = (x + y) (x + y)^2$

$= (x + y)(x^2 + 2xy + y^2)$

$= x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2)$

$= x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3$

$= x^3 + 3x^2y + 3xy^2 + y^3$

$= x^3 + y^3 + 3xy(x + y)$

The RHS side is called the expanded form

**Identity (VII)**

$(x – y)^3 = x^3 – y^3 – 3xy (x – y) $**Derivation:**

This can be easily obtained by y=-y in identity (VI)

**Identity (VIII)**

$x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)$**Derivation:**

\begin{align*}

(x + y + z)&(x^2 + y^2 + z^2 – xy – yz – zx)\\&=x(x^2 + y^2 + z^2 – xy – yz – zx) + y(x^2 + y^2 + z^2 – xy – yz – zx) \\&+ z(x^2 + y^2 + z^2 – xy – yz – zx)\\

&=x^3 + xy^2 + xz^2 – x^2y – xyz – zx^2 + yx^2+ y^3 + yz^2 – xy^2 – y^2z – xyz \\&+ x^2z + y^2z + z^3 – xyz – yz^2 – xz^2\\

&= x^3 + y^3 + z^3 – 3xyz \end{align*}

**Identity (IX)**

$(a + b + c)^3 = a^3+ b^3 + c^3 + 3a^2 b + 3a^2c + 3b^2c +3b^2a +3c^2a +3c^2a+6abc$**Derivation:**

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

Now from identity (V)

\begin{align*}

=&(a+b+c)(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc)\\

=&a( a^2 + b^2 + c^2 + 2ab + 2ac + 2bc )+ b( a^2 + b^2 + c^2 + 2ab + 2ac + 2bc ) + \\&c( a^2 + b^2 + c^2 + 2ab + 2ac + 2bc )\\

= &a^3+ b^3 + c^3 + 3a^2 b + 3a^2c + 3b^2c +3b^2a +3c^2a +3c^2a+6abc

\end{align*}

**Identity (X)**

$x^3 + y^3 = (x + y) (x^2 – xy + z^2 ) $**Derivation:**

From identity (VI)

$(x + y)^3 = x^3 + y^3 + 3xy (x + y)$

$x^3 + y^3 = (x+y)^3 – 3xy (x + y)$

$ x^3 + y^3 = (x+y) [(x+y)^2 – 3xy]$

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

**Identity (XI)**

$x^3 – y^3 = (x – y) (x^2+ xy + y^2 ) $**Derivation:**

From Identity (VII)

$(x – y)^3 = x^3 – y^3 – 3xy (x – y) $

$x^3 – y^3 = (x-y)^3 + 3xy(x-y)$

$x^3 – y^3 = (x-y) [(x-y)^2 +3xy]$

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

**Identity (XII)**

If x + y +z =0, then $x^3+ y^3 + z^3= 3 xyz$**Derivation:**

From Identity (VIII)

$x^3 + y^3 + z^3 – 3xyz = (x + y + z)(x^2 + y^2 + z^2 – xy – yz – zx)$

Now if x + y +z =0

$ x^3 + y^3 + z^3 – 3xyz = 0$

$ x^3+ y^3 + z^3= 3 xyz$

**Identity (XIII)**

$x^2 + y^2 + z^2 – xy – yz – zx= \frac {1}{2} [ (x-y)^2 + (y-z)^2 + (x-z)^2]$

### Solved Examples

**(1) Expand **$(3a + 4b + c)^2$

**Solution **

Comparing the given expression with $(x + y + z)^2$, we find that

x = 3a, y = 4b and z = c.

$(3a + 4b + c)^2 = (3a)^2 + (4b)^2 + (c)^2 + 2(3a)(4b) + 2(4b)(c) + 2(c)(3a)$

$= 9a^2 + 16b^2 + c^2 + 24ab + 8bc + 6a$

**(2) Factorise **$27x^3 + y^3 + 27z^3 – 27xyz$

**Solution**

Here, we have

$27x^3 + y^3 + 27z^3 – 27xyz$

$= (3x)3 + y^3 + (3z)^3 – 3(3x)(y)(3z)$

$= (3x + y + 3z)[(3x)^2 + y^2 + (3z)^2 – (3x)(y) – (y)(3z) – (3x)(3z)]$

$= (3x + y + 3z) (9x^2 + y^2 + 9z^2 – 2xy – 3yz – 6xz)$

**(3) Expand** $(3p + 4q)^3$**Solution**

Comparing the given expression with $(x + y)^3$, we find that

x = 3p and y = 4q.

So, using Identity VI, we have:

$(3p + 4q)^3 = (3p)^3 + (4q)^3 + 3(3p)(4q)(3p + 4q)$

$= 27p^3 + 64q^3 + 108p^2q + 144pq^2$

## Summary of All the Algebraic Identities

**Also Reads**

- Learn about the Algebraic Expressions and Identities Notes for Class 8
- BODMAS Rule
- Rules of Surds
- https://en.wikipedia.org/wiki/Identity_(mathematics)