Question

Factorise : $(a-b{)}^3+(b-c{)}^3+(c-a{)}^3$.

• A. $3a^2(b-c)+3b^2(c-a)+3c^2(a-b)$
• B. $a^2(c-b)+b^2(a-c)+c^2(b-c)$
• C. $3a^2(c-b)+3b^2(a-c)+3c^2(b-a)$
• D. $3a(c-b)+3b(a-c)+3c(b-c)$

Answer 1

Answer: (C)

We know, $(a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2$

$⟹$ $(a-b{)}^3=a^3-b^3-3ab(a-b)$.

Then,

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

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

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

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

Therefore, option $C$ is correct.