Question

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

• A. $-98$
• B. $98$
• C. $96$
• D. $-96$

Answer 1

The correct option is B $98$

$f\left(x\right)=4x^3-7x^2+1$
$f\left(\frac{1}{x}\right)=x^3-7x+4$
Lets take $\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma }$ as $a,b,c$
Multiply both sides by $x$, then
$x^4-7x^2+4x=0$
$a^4=7a^2-4a$
$b^4=7b^2-4b$
$c^4=7c^2-4c$
$a^4+b^4+c^4=7\left[\right(a+b+c{)}^2-2(ab+bc+ca)]-4(a+b+c)$
$a^4+b^4+c^4=7(0-2(-7\left)\right)$
$a^4+b^4+c^4=98$