Question

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

• A. 0
• B. 1
• C. -1
• D. Not defined

Answer 1

Answer: (B)

1

given that, $a+b+c=0$

So, $b+c=-a$, $c+a=-b$, $a+b=-c$

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

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

So, $\frac{3abc}{3abc}=1$
Answer (B) 1