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# Factorisation using cubic identities

Algebra: Factorisation using cubic identities

Factorisation using cubic identities

The cubic identities are

(i) (a + b)3 = a3 + 3a2b + 3ab2 + b3

(ii) (a b)3 = a3 3a2b + 3ab2 b3

Note

8a 3 = 2 × 2 × 2 × a3

= 23 a3 = (2a)3

I. Factorise using the identity (a + b)3 = a3 + 3a2b + 3ab2 + b3

Example 3.26

Factorise: x3 + 15x2 + 75x + 125

Solution:

x3 + 15x2 + 75x + 125

This can be written as x3 + 15x2 + 75x + 53

Comparing with a 3 + 3a2b + 3ab 2 + b3 = (a + b)3

we get a = x , b = 5

The given expression can be expressed as

(x)3 + 3(x)2 (5) + 3(x) (5)2 + (5)3 = (x + 5)3

= (x + 5) (x + 5) (x + 5) are the three factors.

Note

Perfect cube numbers

A number which can be written in the form of x × x × x is called perfect cube number

Examples

8=2×2×2=23

27=3×3×3=33

125=5×5×5=53

Here 8, 27, 125 are some of perfect cube numbers

II. Factorise using the identity (ab)3 = a3 – 3a2b + 3ab2b3

Example 3.27

Factorise: 8p3 – 12p2q + 6pq2q3

Solution:

Given 8p3 – 12p2q + 6pq2q3

This can be written as

(2p)3 – 12p2q + 6pq2 – (q)3

Comparing this with a 3 3a2b + 3ab 2 b3 = (a b)3 we get a = 2p , b = q

The given expression can be expressed as

(2p)3 – 3(2p)2(q) + 3(2p)(q)2 – (q)3 = (2pq)3

= (2pq), (2pq), (2pq) are the three factors.

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