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(B) Given the equation $x^{2} + 3^{1/4}x + 3^{1/2} = 0$.Since $\alpha$ is a root,$\alpha^{2} + 3^{1/2} = -3^{1/4}\alpha$.Squaring both sides: $(\alpha

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$52 	imes 3^{24}$

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(B) Given the equation $x^{2} + 3^{1/4}x + 3^{1/2} = 0$.Since $\alpha$ is a root,$\alpha^{2} + 3^{1/2} = -3^{1/4}\alpha$.Squaring both sides: $(\alpha^{2} + 3^{1/2})^{2} = (3^{1/4}\alpha)^{2} = 3^{1/2}\alpha^{2}$.$\alpha^{4} + 2 \cdot 3^{1/2}\alpha^{2} + 3 = 3^{1/2}\alpha^{2}$.$\alpha^{4} + 3^{1/2}\alpha^{2} + 3 = 0$.Multiply by $(\alpha^{2} - 3^{1/2})$: $(\alpha^{2} - 3^{1/2})(\alpha^{4} + 3^{1/2}\alpha^{2} + 3) = 0$.This is the form $(a-b)(a^{2}+ab+b^{2}) = a^{3}-b^{3}$,so $\alpha^{6} - (3^{1/2})^{3} = 0$.$\alpha^{6} = 3^{3/2} = 3 \sqrt{3}$.Then $\alpha^{12} = (3 \sqrt{3})^{2} = 9 	imes 3 = 27 = 3^{3}$.Since $\alpha^{12} = 27$,then $\alpha^{96} = (\alpha^{12})^{8} = (27)^{8} = (3^{3})^{8} = 3^{24}$.Similarly,$\beta^{12} = 27$ and $\beta^{96} = 3^{24}$.The expression is $\alpha^{96}(\alpha^{12}-1) + \beta^{96}(\beta^{12}-1) = 3^{24}(27-1) + 3^{24}(27-1)$.$= 3^{24}(26) + 3^{24}(26) = 2 	imes 26 	imes 3^{24} = 52 	imes 3^{24}$.

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:

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