Question

Let $\alpha, \beta, \gamma$ are the roots of the cubic equation $a_0 x^3+3 a_1 x^2+3 a_2 x+a_3=0 \quad\left(a_0 \neq 0\right)$. Then the value of $(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2$ equals

• A. $\frac{18\left( a _2^2- a _0 a _1\right)}{ a _0^2}$
• B. $\frac{18\left( a _2^2+ a _0 a _1\right)}{ a _0^2}$
• C. $\frac{18\left( a _0^2- a _1 a _2\right)}{ a _0^2}$
• D. $\frac{18\left( a _1^2- a _0 a _2\right)}{ a _0^2}$

Answer 1

Answer: (D)

$\Sigma \alpha=\alpha+\beta+\gamma=\frac{-3 a_1}{a_0} $
$\Sigma \alpha \beta=\alpha \beta+\beta \gamma+\gamma \alpha=\frac{3 a_2}{a_0}$
$\text { Now, }(\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2=2\left(\alpha^2+\beta^2+\gamma^2\right)-2 \Sigma \alpha \beta=2\left((\alpha+\beta+\gamma)^2-2 \Sigma \alpha \beta\right)-2 \Sigma \alpha \beta$
$=2(\alpha+\beta+\gamma)^2-6 \Sigma \alpha \beta=2\left(\frac{-3 a_1}{a_0}\right)^2-6\left(\frac{3 a_2}{a_0}\right)=\frac{18 a_1^2}{a_0^2}-\frac{18 a_2}{a_0}=\frac{18\left(a_1^2-a_0 a_2\right)}{a_0^2}$