Q. 74.6( 11 Votes )

# Let us factorise the following algebraic expressions:

a^{6} + 32a^{3} – 64

Answer :

a^{6} + 32a^{3} – 64

⇒ a^{6} + 8a^{3} – 64 + 24a^{3}

⇒ (a^{2})^{3} + (2a) ^{3} + (- 4)^{3} - 3×(a^{2})×(2a)×(- 4) …Equation (i)

We use the identity

a^{3} + b^{3} + c^{3} - 3abc = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca)

Using the above identity in Equation (i) we get

⇒ (a^{2})^{3} + (2a) ^{3} + (- 4)^{3} - 3×(a^{2})×(2a)×(- 4) = (a^{2} + 2a - 4) (a^{4} + 4a^{2} + 16 - 2a^{3} + 8a + 4a^{2})

⇒ (a^{2} + 2a - 4) (a^{4} + 8a^{2} + 16 - 2a^{3} + 8a)

(a^{2} + 2a - 4) (a^{4} - 2a^{3} + 8a^{2} + 8a + 16)

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