Question

lf $\alpha ,\beta ,\gamma$ are the roots of $x^3+px+q=0$, then $\frac{{\alpha }^5+{\beta }^5+{\gamma }^5}{5}=$

• A. $\frac{{\alpha }^3+{\beta }^3+{\gamma }^3}{3}\cdot \frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{2}$
• B. $\frac{{\alpha }^3+{\beta }^3+{\gamma }^3}{3}+\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{2}$
• C. $\frac{{\alpha }^3+{\beta }^3+{\gamma }^3}{3}-\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{2}$
• D. $\frac{{\alpha }^3+{\beta }^3+{\gamma }^3}{3}÷\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{2}$

Answer 1

The correct option is A $\frac{{\alpha }^3+{\beta }^3+{\gamma }^3}{3}\cdot \frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{2}$

Given, $x^3+px+q=0,\alpha ,\beta ,\gamma$ are roots of the equation

$\Rightarrow \alpha +\beta +\gamma =0\left(S_1\right)$

$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =p\left(S_2\right)$

$\alpha \beta \gamma =-q\left(S_3\right)$

Say $P_n={\alpha }^n+{\beta }^n+{\gamma }^n$

We have newton's identities

$P_1=S_1=0$

$P_2=S_1{P}_1-2S_2=-2S_2$

$P_3=S_1{P}_2-S_2{P}_1+3S_3=3S_3$

$P_4=S_1{P}_3-S_2{P}_2+S_3{P}_1=-S_2{P}_2=2{S_2}^2$

$P_5=S_1{P}_4-S_2{P}_3+S_3{P}_2=-3{S_{2}S}_3-2{S_{2}S}_3=-5{S_{2}S}_3$

$\frac{P_5}{5}=-{S_{2}S}_3=\left(\frac{P_2}{2}\right)\left(\frac{P_3}{3}\right)=\left(\frac{{\alpha }^2+{\beta }^2+{\gamma }^2}{2}\right)\left(\frac{{\alpha }^3+{\beta }^3+{\gamma }^3}{3}\right)$