Question

factorise 8 a cube minus b cube + 64 C Cube + 24 ABC

Answer 1

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


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

Answer 2

Given polynomial :
${8a}^{3} - {b}^{3} + 64 {c}^{3} + 24abc$
Identity to be used :

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

${8a}^{3} - {b}^{3} + {64c}^{3} + 24abc \\ {(2a)}^{3} + {( - b)}^{3} + {(4c)}^{3} - 3(2a)( - b)(4c)$
(2a-b+4c)[(2a)²+(-b)²+(4c)²-(2a)(-b)-(-b)(4c)-(4c)(2a)]

= (2a-b+4c)(4a²+b²+16c²+2ab+4bc-8ca)


#Be Brainly !!