Question

If $a+b+c=0$ , then $\frac{(b+c{)}^2}{3bc}+\frac{(c+a{)}^2}{3ac}+\frac{(a+b{)}^2}{3ab}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

The expression is

$\frac{(b+c{)}^2}{3bc}+\frac{(c+a{)}^2}{3ac}+\frac{(a+b{)}^2}{3ab}$

If $a+b+c=0$

then

$b+c=-a$

$c+a=-b$

$a+b=-c$

Substitute these values in the expression,

$\frac{(-a{)}^2}{3bc}+\frac{(-b{)}^2}{3ac}+\frac{(-c{)}^2}{3ab}$

$=\frac{a^2}{3bc}+\frac{b^2}{3ac}+\frac{c^2}{3ab}$

$=\frac{a^3+b^3+c^3}{3abc}$

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

$=\frac{\left(0\right)(a^2+b^2+c^2-ab-bc-ac)+3abc}{3abc}$

$=\frac{3abc}{3abc}$

$=1$