JEE Main 2021 — Mathematics Algebra

We have,${x}^{2}+{(3)}^{\frac{1}{4}}x+{3}^{\frac{1}{2}}=0$x2+(3)41​x+321​=0

Since, $\alpha$α is the root, therefore $({\alpha }^{2}+\sqrt{3})=-{(3)}^{\frac{1}{4}}\cdot \alpha$(α2+3​)=−(3)41​⋅α

On squaring, both sides we get

$$({\alpha }^{4}+2\sqrt{3}{\alpha }^{2}+3)=\sqrt{3}{\alpha }^{2}$$(α4+23​α2+3)=3​α2

$$\Rightarrow {\alpha }^{4}+3=-\sqrt{3}{\alpha }^{2}$$⇒α4+3=−3​α2

On squaring, both sides we get

$${\alpha }^{8}+6{\alpha }^{4}+9=3{\alpha }^{4}$$α8+6α4+9=3α4

$$\Rightarrow {\alpha }^{8}=-9-3{\alpha }^{4}$$⇒α8=−9−3α4

Multiply by ${\alpha }^{4}$α4, we get

$${\alpha }^{12}=-9{\alpha }^{4}-3{\alpha }^{8}$$α12=−9α4−3α8

$$\therefore {\alpha }^{12}=-9{\alpha }^{4}-3(-9-3{\alpha }^{4})$$∴α12=−9α4−3(−9−3α4)

$$\Rightarrow {\alpha }^{12}=27$$⇒α12=27

$$\Rightarrow {({\alpha }^{12})}^{8}={(27)}^{8}$$⇒(α12)8=(27)8

$$\Rightarrow {\alpha }^{96}={(3)}^{24}$$⇒α96=(3)24

Similarly,

$${\beta }^{96}={(3)}^{24}$$β96=(3)24

$$\therefore {\alpha }^{96}({\alpha }^{12}-1)+{\beta }^{96}({\beta }^{12}-1)=(3){}^{24}+(27+27-2)={(3)}^{24}\times 52$$∴α96(α12−1)+β96(β12−1)=(3)24+(27+27−2)=(3)24×52