Question

If $a,b,c$ are all non-zero and $a+b+c=0$, then $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}=$

• A. Not equal to $3$
• B. Equal to zero
• C. $3$
• D. Insufficient data

Answer 1

The correct option is C $3$

$\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}$

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

$=\frac{a^3+b^3+c^3}{abc}$ ...... $\left(i\right)$

Now,

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

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

$(a+b+c{)}^3=a^3+b^3+c^3+3a^{2}b+3a^{2}c+3a{b}^2+3b^{2}c+3a{c}^2+3b{c}^2+6abc$

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

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

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

$(a+b+c{)}^3=(a^3+b^3+c^3)+3[(a+b+c)(ab+ac+bc)-abc]$

Now,

$(a+b+c)=0$

Thus, $0=(a^3+b^3+c^3)+3\left[\right(a+b+c\left)\right(ab+ac+bc)-abc]$

$=(a^3+b^3+c^3)+3\left[\right(0\left)\right(ab+ac+bc)-abc]$

$=(a^3+b^3+c^3)+3\left[\right(0)-abc]$

$=(a^3+b^3+c^3)-3abc$

$\Rightarrow (a^3+b^3+c^3)=3abc$

Putting the value of $(a^3+b^3+c^3)$ in $\left(i\right)$ we get,

$\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}$

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

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

$=3$