Question

find the factor of (2a-b)^3+(b-2c)^3+8(c-a)^3​

Answer 1

» Question -

find factor of (2a-b)³ + (b-2c)³ +8(c-a)³

» Solution -

8(c-a)³ = 2³ (c-a)³ = {2(c-a)} ³ = (2c-2a)³

therefore,

(2a-b)³ + (b-2c)³ +8(c-a)³

= (2a-b)³ + (b-2c)³ +(2c-2a)³

We know that,

${\pink{\boxed{x^3+y^3+z^3 = 3xyz, if\:x+y+z = 0}}}$

in our case,

x = 2a-b

y = b-2c

z = 2c-2a

x+y+z =

2a-b+b-2c+2c-2a

= 0(all terms get cancelled)

hence, we can use the identity.

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

= ${\purple{\boxed{6(2a-b)(b-2c)(c-a)}}}$