Question

Factorise 8a^3-b^3-64c^3-24abc

Answer 1

8a³-b³-64c³-24abc

= 2³a³-b³-4³c³-3×2a×b×4c

= (2a)³-b³-(4c)³

Identify to be used :

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


Now ,
(2a)³-b³-(4c)³ = (2a-b+4c)(4a²+b²+16c²+2ab+4bc-8ac)

Thanks !!

Answer 2

Solution

8a³ -b³ -64c³ -24abc

(2)³ - (b) ³ -(4)³ -3(2a)(-b) (-4)

Factorise given below

$<b>$ we use the identity

a³ + b³ +c³ -3(a)(b)(c) =(a+b+c)(a²+b².+c² -ab-bc-ca)

( 2a - b + 4c) = ( 4a² -b² +16c² 2ab + 4bc - 8ac)