Question

The factors of $(2a-b{)}^3+(b-2c{)}^3+8(c-a{)}^3$ are :

• A. $(2a-b)(b-2c)(c-a)$
• B. $3(2a-b)(b-2c)(c-a)$
• C. $6(2a-b)(b-2c)(c-a)$
• D. $2a\times b\times 2c$

Answer 1

Answer: (A), (B), (C)

$(2a-b)(b-2c)(c-a)$
$3(2a-b)(b-2c)(c-a)$
$6(2a-b)(b-2c)(c-a)$

$a^3+b^3{+c}^3-3abc=(a+b+c)(a^2+b^2{+c}^2-ab-bc-ca)$

So, if $a+b+c=0$, $a^3+b^3{+c}^3=3abc$

The given expression can be written as

${(2a-b)}^3+{(b-2c)}^3+{(2c-2a)}^3$

We can see here that $2a-b+b-2c+2c-2a=0$.

Hence, following the above concept

${(2a-b)}^3+{(b-2c)}^3+{(2c-2a)}^3=6(2a-b)(b-2c)(c-a)$

Hence, options A, B, C are correct.