Algebra: Factorisation using cubic identities

**Factorisation
using cubic identities**

The cubic
identities are

(i) (*a* + *b*)^{3}
=
*a*^{3} +
3*a*^{2}*b* + 3*ab*^{2} +
*b*^{3}

(ii) (*a* − *b*)^{3}
=
*a*^{3} −
3*a*^{2}*b* + 3*ab*^{2} −
*b*^{3}

**Note**

8*a* ^{3} = 2 × 2 × 2 × *a*^{3}

= 2^{3} *a*^{3} = (2*a*)^{3}

** **

__I. Factorise
using the identity ( a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2}
+ b^{3}__

**Example 3.26**

Factorise:
*x*^{3} + 15*x*^{2} + 75*x* + 125

*Solution:*

*x*^{3}* *+ 15*x*^{2}* *+ 75*x
*+ 125

This can
be written as *x*^{3} + 15*x*^{2} + 75*x* + 5^{3}

Comparing
with *a* ^{3} +
3*a*^{2}*b* + 3*ab* ^{2} +
*b*^{3} = (*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=2^{3}

27=3×3×3=3^{3}

125=5×5×5=5^{3}

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

** **

__II. Factorise
using the identity ( a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2}
– b^{3}__

**Example 3.27**

Factorise:
8*p*^{3} – 12*p*^{2}*q* + 6*pq ^{2}* –

*Solution:*

Given 8*p*^{3} – 12*p*^{2}*q* + 6*pq ^{2}* –

This can
be written as

(2p)^{3}
– 12*p*^{2}*q* + 6*pq ^{2}* – (

Comparing
this with *a* ^{3} −
3*a*^{2}*b* + 3*ab* ^{2} −
*b*^{3} = (*a* − *b*)^{3} we get *a* = 2*p* ,
*b* = *q*

The given
expression can be expressed as

(2p)^{3}
– 3(2*p*)^{2}*(q)* + 3(2p)*(q)*^{2} – (*q*)^{3}
= (2*p* –*q*)^{3 }

= (2*p* –*q*),
(2*p* –*q*), (2*p* –*q*) are the three factors.

Tags : Algebra | Chapter 3 | 8th Maths , 8th Maths : Chapter 3 : Algebra

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