Question

If $\alpha ,\beta ,\gamma$ are the roots of the cubic $210x^3+4x^2+1=0$, then the value of $({\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2})$ is equal to

• A. $8$
• B. $-8$
• C. $4$
• D. $-4$

Answer 1

The correct option is B $-8$

Given cubic equation is $210x^3+4x^2+1=0$

$\alpha ,\beta ,\gamma$ are the roots of the cubic equation

Here, $\alpha +\beta +\gamma =-\frac{4}{210}$

$\alpha \beta +\beta \gamma +\alpha \gamma =0$

$\alpha \beta \gamma =-\frac{1}{210}$

So, $(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2})=\left(\frac{(\alpha \beta {)}^2+(\beta \gamma {)}^2+(\alpha \gamma {)}^2}{(\alpha \beta \gamma {)}^2}\right)=\frac{(\alpha \beta +\beta \gamma +\alpha \gamma {)}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )}{(\alpha \beta \gamma {)}^2}$

On simplifying and substituting the relations between roots and coefficients of the polynomial,

$(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2})=-2\times \frac{-4}{210}\frac{210}{-1}=(-8)$

So, Option B is the correct answer.