# Algebraic Identities

Complete above steps to enable this content

From your earlier classes, you may recall that an algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. You have studied the following algebraic identities in earlier classes:

**Identity I :**

**Identity II :**

**Identity III :**

**Identity IV :** (x + a) (x + b) =

**Identity V :**

**Identity VI :**

**Identity VII :**

**Identity VIII :**

You must have also used some of these algebraic identities to factorise the algebraic expressions. You can also see their utility in computations.

Complete above steps to enable this content

**Example 10**

**Find the following products using appropriate identities:**

**(i) (x + 3) (x + 3)**

- We have the equation, Here we can use Identity I is :
- using Identity I
=x + y 2 +2xy+x 2 Putting y = 3 in it. we have to apply the identity I and Y value in our producty 2 - calcualte the terms and we get the equation is
+x 2 x +

**(ii) (x – 3) (x + 5)**

- We have the equation, Here we can use Identity IV is :
- using Identity IV (x + a) (x + b) =
+ (a + b)x + ab apply (x-3)(x+5) into the identityx 2 - calculate the terms and we get the equation is
+x 2 x -

Complete above steps to enable this content

**Example 11**

**Evaluate 105 × 106 without multiplying directly.**

- We have expression is
- expand the expression with first term 100 =
+ and the seconde term is 106 = + - using Identity IV ((x + a) (x + b) =
+ (a + b)x + ab)we haveto apply the IV identity in our expressionx 2 - calcuate and add the terms
- We get the answer is

You have seen some uses of the identities listed above in finding the product of some given expressions. These identities are useful in factorisation of algebraic expressions also, as you can see in the following examples.

**Example 12**

**Factorise:**

- Now comparing it with Identity III We get
- So the find the factors of 25 =
, 4= and 9 = - Therefore the expression is

So far, all our identities involved products of binomials.

Complete above steps to enable this content

**Example 13**

**Let us now extend the Identity I to a trinomial x + y + z. We shall compute (x + y + z)^2 by using Identity I.**

- We have the expression, Let x + y = t. Then,
- Using Identity I (
=x + y 2 +2xy+x 2 ), so the expression isy 2 + 2t 2 + z 2 - Substituting the value of t is
+2x + y 2 z + z 2 - Using Identity I
- Rearranging the terms

**Remark :** We call the right hand side expression the expanded form of the left hand side expression. Note that the expansion of

**Example 14**

**Write **

- Comparing the given expression with
we find that x = 3a, y = 4b and z = 5c.x + y + z 2 - Therefore, using Identity V, we have
- expand the expression
- We get the expression is
+a 2 +b 2 +c 2 ab+ bc+ ac

Complete above steps to enable this content

**Example 15**

**Write the following cubes in the expanded form:**

- Comparing the given expression with
we find that x = 3a, y = 4b.x + y 3 - Therefore, using Identity VI, we have
- expand the expression
- We get the expression is
+a 3 +b 3 b+a 2 a b 2

**Example 16**

**Factorise : 8**

- Here, we have
- expand the expression
- Substituting the expression with factor
- Hence we get the expression is