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(D) Given equation is $x^2-2 \sqrt{3} x+4=0$ with roots $\alpha$ and $\beta$.From the properties of roots,$\alpha+\beta = 2 \sqrt{3}$ and $\alpha \bet

Vedclass · Accepted Answer

-$128$

Vedclass · Answer

(D) Given equation is $x^2-2 \sqrt{3} x+4=0$ with roots $\alpha$ and $\beta$.From the properties of roots,$\alpha+\beta = 2 \sqrt{3}$ and $\alpha \beta = 4$.First,calculate $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (2 \sqrt{3})^2 - 2(4) = 12 - 8 = 4$.Now,we use the identity $\alpha^6+\beta^6 = (\alpha^2)^3 + (\beta^2)^3 = (\alpha^2+\beta^2)((\alpha^2+\beta^2)^2 - 3\alpha^2\beta^2)$.Substituting the values: $\alpha^6+\beta^6 = (4)((4)^2 - 3(4)^2) = 4(16 - 3(16)) = 4(16 - 48) = 4(-32) = -128$.Thus,the correct option is $D$.

If $\alpha, \beta$ are the roots of the equation $x^2-2 \sqrt{3} x+4=0$,then $\alpha^6+\beta^6=$

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