Question

If $\alpha$ and $\beta$ are the distinct roots of the equation $x^2+(3{)}^{1/4}x+3^{1/2}=0,$ then the value of ${\alpha }^{96}({\alpha }^{12}-1)+{\beta }^{96}({\beta }^{12}-1)$ is equal to

• A. $28\times 3^{25}$
• B. $56\times 3^{24}$
• C. $52\times 3^{24}$
• D. $56\times 3^{25}$

Answer 1

The correct option is C $52\times 3^{24}$

Given equation is $x^2+(3{)}^{1/4}x+3^{1/2}=0$
$\therefore$ $x=\frac{-3^{1/4}\pm \sqrt{3^{1/2}-4\cdot 3^{1/2}}}{2}$
$=\frac{3^{1/4}(-1\pm \sqrt{3}i)}{2}=3^{1/4}\omega$ or $3^{1/4}{\omega }^2$

${\alpha }^{108}+{\beta }^{108}-({\alpha }^{96}+{\beta }^{96})=3^{108/4}({\omega }^{108}+{\omega }^{216})-3^{96/4}({\omega }^{96}+{\omega }^{192})=3^{27}\cdot 2-3^{24}\cdot 2=3^{24}\cdot 2(27-1)=52\times 3^{24}$