Factorise:a^{3}-8 b^{3}-64 c^{3}-24 a b c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We use the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.Given expression: $a^{3}-8b^{3}-64c^{3}-24abc$This c

Vedclass · Accepted Answer

$(a-2b-4c)(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ca)$

Vedclass · Answer

(A) We use the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.Given expression: $a^{3}-8b^{3}-64c^{3}-24abc$This can be rewritten as: $a^{3}+(-2b)^{3}+(-4c)^{3}-3(a)(-2b)(-4c)$.Here,$x=a$,$y=-2b$,and $z=-4c$.Applying the identity:$= (a-2b-4c)((a)^{2}+(-2b)^{2}+(-4c)^{2}-(a)(-2b)-(-2b)(-4c)-(-4c)(a))$$= (a-2b-4c)(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ca)$.

Factorise:
$a^{3}-8 b^{3}-64 c^{3}-24 a b c$

Similar Questions

The coefficient of $x$ in the expansion of $(x+3)^{3}$ is

Evaluate the following product without multiplying directly:
$93 \times 95$

Factorise the following: $27-125 a^{3}-135 a+225 a^{2}$

Multiply $x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z$ by $(-z+x-2 y)$.

Factorise the following expression: $8 x^{3}+27 y^{3}+36 x^{2} y+54 x y^{2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Factorise:$a^{3}-8 b^{3}-64 c^{3}-24 a b c$