Question

a^3-8b^3+64c² +24 abc​

Answer 1

Step-by-step explanation:

Given that:

$a^3-8b^3+64c^3+24abc$

To find: Factorize the expression

Solution: Identity used

$\bold{x^3+y^3+z^3-3xyz}\\\bold{=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}$

Convert the expression

$a^3-8b^3+64c^3+24abc$

So that it can be expanded

$a^3+(-2b)^3+(4c)^3-3(a)(-2b)(4c)\\\\=(a-2b+4c)[a^2+(-2b)^2+(4c)^2-(a)(-2b)-(-2b)(4c)-(a)(4c)]$

$\bold{a^3-8b^3+64c^3+24abc}\\\\\bold{=(a-2b+4c)[a^2+4b^2+16c^2+2ab+8bc-4ac]}$

Hope it helps you.