Question

factorise 8 a cube minus b cube + 64 C + 24ab. c




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Answer 1

Answer:

$\Large{\underline{\underline{\bf{QuEsTiOn:-}}}}$

Factorise 8 a cube minus b cube + 64 C + 24 ABC

$\Large{\boxed{\underline{\overline{\mathfrak{\star \: AnSwer :- \: \star}}}}}$

We have to factorise:

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

For it, we will use the following identity:

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

We can factorize using the identity as follows:

$8a^3-b^3+64c^3+24abc \\ \\ = (2a)^3 + (-b)^3 + (4c)^3 -3(2a)(-b)(4c) \\ \\ = (2a-b+4c)((2a)^2+(-b)^2+(4c)^2-(2a)(-b)-(-b)(4c)-(4c)(2a)) \\ \\ = (2a-b+4c)(4a^2+b^2+16c^2+2ab+4bc-8ca)$

Thus, The factorized form is

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

Hope it helps uhh

Answer 2

Answer:

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