Question

Choose the correct answer from the alternatives given.
Value of the expression
$\frac{(a-b{)}^2}{(b-c)(c-a)}+\frac{(b-c{)}^2}{(a-b)(c-a)}+\frac{(c-a{)}^2}{(a-b)(b-c)}$ is.

• A. $0$
• B. $3$
• C. $\frac{1}{3}$
• D. $2$

Answer 1

Answer: (B)

$3$

$\frac{(a-b{)}^2}{(b-c)(c-a)}+\frac{(b-c{)}^2}{(a-b)(c-a)}+\frac{(c-a{)}^2}{(a-b)(b-c)}$
$\frac{(a-b{)}^3+(b-c{)}^3+(c-a{)}^3}{(a-b)(b-c)(c-a)}$
Since.$(a-b)+(b-c)+(c-a)=a-b+b-c+c-a=0$
$\therefore (a-b{)}^3+(b-c{)}^3+(c-a{)}^3$ =$3(a-b)(b-c)(c-a)]$

Hence, $\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}=3$