Question

factorise 8 a^3- b ^3+ 64c^3 + 24 ABC

Answer 1

Hi friend!!

Given,

8a³-b³+64c³+24abc

= (2a)³+(-b)³+(4c)³-3(2a)(-b)(4c)

we know that x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)

→(2a-b+4c)(4a²+b²+16c²+2ab+4bc-8ac)

I hope this will help you ;)

Answer 2

8a^3-b^3+64c^3+24abc
(2a)^3-(b)^3+(4c)^3 - (3×2a×-b×4c)

using identity (a)^3+(b)^3+(c)^3 -3abc = (a+b+c)(a^2 +b^2 +c^2-ab-bc-ca)

answer is [2a+(-b)+4c][(2a)^2+(-b)^2+(4c)^2 -2a×(-b) - (-b)×4c - 2a×4c]
(2a-b+4c)(4a^2+b^2+16c^2 + 2ab + 4bc - 8ac