Algebraic Identities
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) =
You must have also used some of these algebraic identities to factorise the algebraic expressions. You can also see their utility in computations.
Example 11
Find the following products using appropriate identities:
(i)
- We have the equation, Here we can use Identity I is :
- using Identity I
x + y 2 x 2 y 2 - Calculating the terms and we get the equation:
x 2 x + - We have found the solution.
(ii) (x – 3) (x + 5)
- We have the equationas shown above. Here we can use Identity IV.
- Using Identity IV (x + a) (x + b) =
x 2 - Calculating the terms and we get the equation:
x 2 x - - We have found the solution.
Example 12
Evaluate 105 × 106 without multiplying directly.
- Taking the product, express each term as a sum of two numbers.
- Expanding the expression we get: First term 105 = 100 +
and the second term 106 = 100 + - Using Identity IV
x + a x + b x 2 + a + b x + ab we have to apply the IV identity in the expression. - Calculating and adding the terms.
- We get the answer as
. - We have found the answer.
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 13: Factorise:
(i)
- Now comparing it with Identity I we get:
-
49 a 2 , 25 b 2 , 70 ab = - Thus,
x =and y = - Therefore the expression is:
49 a 2 + 70 ab + 25 b 2 - We have found the answer.
(ii)
- Now comparing it with Identity III we get
- So the find the factors of
25 , 4 and 9 - Therefore the expression is
- We have found the answer.
So far, all our identities involved products of binomials.
Let us now extend the Identity I to a trinomial x + y + z. We shall compute
- We have the expression, Let x + y = t. Then,
- Using Identity I (
x + y 2 x 2 y 2 t 2 + z 2 - Substituting the value of t is
x + y 2 z + z 2 - Using Identity I
- Rearranging the terms
- We have found the answer.
Identity V :
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
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 - We have found the answer.
Example 15 :
Example 16:
So far, we have dealt with identities involving second degree terms. Now let us extend Identity I to compute
=
=
=
This gives us:
Identity VI :
Identity VII :
Example 17
Write the following cubes in the expanded form:
(i)
- Comparing the given expression with
x + y 3 - Therefore, using Identity VI, we have
- expand the expression
- We get the expression is
a 3 b 3 a 2 b +ab 2 - We have found the answer.
(ii)
- Comparing the given expression with
x + y 3 - Therefore, using Identity VII, we have
- Expand the expression
5 p − 3 q 3 5 p 3 − 3 q 3 − 3 5 p 3 q 5 p − 3 q - We get the expression is
+ – + - We have found the answer.
Example 18 :
Example 19
Factorise :
- Here, we have
- expand the expression
- Substituting the expression with factor
- Hence we get the expression is
Identity VIII :
Example 20: