Question

If $a{x}^2+bx+c=0andc{x}^2+bx+a=0$ (a,b,c,$ϵ$,R) have a common non-real root, then

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

• A. $3$
• B. $2$
• C. $5$
• D. $7$

Answer 1

Answer: (A)

$3$

We have,

$a{x}^2+bx+c=0$…… (1)

$c{x}^2+bx+a=0$…… (2)

Have a common real roots.

Then,

Subtract equation (1)- (2) to and we get,

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

$\Rightarrow a{x}^2+c-c{x}^2-a=0$

$\Rightarrow a{x}^2-c{x}^2+c-a=0$

$\Rightarrow x^2(a-c)+c-a=0$

$\Rightarrow x^2(a-c)=a-c$

$\Rightarrow x^2=1$

$\Rightarrow x=\pm 1$

Put $x=\pm 1$ in both equation and we get,

$a+b+c=0$

$a+b+c=0$

Then the value of

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

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

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

$=3$

Hence, this is the answer.