Question

lf $\alpha ,\beta ,\gamma$ are the roots of $x^3-7x^2+14x-8=0$, then the value of $\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}=$

• A. $\frac{273}{256}$
• B. $-\frac{273}{256}$
• C. $\frac{264}{273}$
• D. $-\frac{264}{273}$

Answer 1

The correct option is A $\frac{273}{256}$

As $x=1$ is one of the roots .
Other two roots are $2$ and $4$ .
$\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}=1+\frac{1}{2^4}+\frac{1}{4^4}=\frac{273}{256}$