Question

The value of $\frac{{(a-b)}^3+{(b-c)}^3+{(c-a)}^3}{(a-b)(b-c)(c-a)}$ is

• A. $1$
• B. $3$
• C. $\frac{1}{3}$
• D. zero

Answer 1

The correct option is B $3$

Since $(a-b)+(b-c)+(c-a)=0$
$\therefore$ ${(a-b)}^3+{(b-c)}^3+{(c-a)}^3=3(a-b)(b-c)(c-a)$
$\Rightarrow \frac{{(a-b)}^3+{(b-c)}^3+{(c-a)}^3}{(a-b)(b-c)(c-a)}=\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3$