Question

The equation ${ax}^2+bx+c=0,{bx}^2+cx+a=0$ have a common root then $\frac{a^3+b^3+c^3}{abc}$

• A. 1
• B. 3
• C. 2
• D. 4

Answer 1

The correct option is B 3

We have,

$a{x}^2+bx+c=0$ and $b{x}^2+cx+a=0$ have a common root

Then,

$a{x}^2+bx+c=b{x}^2+cx+a=0$

If put $x=1$ then,

$a+b+c=a+b+c=0$

So,

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

Using formula,

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Then,

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

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

$=\frac{0}{3abc}+3$

$=3$

Hence, this is the answer.