Question

If $\alpha ,\beta$ and $\gamma$ are the three zeroes of the polynomial $p\left(x\right)=x^3-64x-14$, what is the value of ${\alpha }^3+{\beta }^3+{\gamma }^3$?

• A. $36$
• B. $40$
• C. $42$
• D. $64$

Answer 1

The correct option is C $42$

${\alpha }^3+{\beta }^3+{\gamma }^3=(\alpha +\beta +\gamma )({\alpha }^2+{\beta }^2+{\gamma }^2-\gamma \beta -\beta \gamma -\gamma \alpha )+3\alpha \beta \gamma$

$\therefore P\left(x\right)=x^3-64x-14$

$\therefore \alpha +\beta +\gamma =\frac{-b}{a}=\frac{0}{1}=0$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}=-64$

$\therefore \alpha \beta \gamma =\frac{-d}{a}=14$

$\therefore {\beta }^3+{\beta }^3+{\gamma }^3=\left(0\right)({\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\gamma \alpha )+3\left(14\right)$

$\therefore {\alpha }^3+{\beta }^3+{\gamma }^3=42$