25 If a and B are roots of the equation, 2 - 412 kx + 2e4in k – 1 = 0 for some k, and Q2 + B2 = 66, then a3 + b3 is equal to : [JEE Main Online - 2014] (A) 248 12 (B) 280 V2 (C) -32 V2 (D) –280 V2

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$$\begin{gathered} = {\mathrm{x}}^2-4\sqrt{2}\mathrm{xn}+\left(2{\mathrm{x}}^4-1\right)=0 \\ \mathrm{\alpha }+\mathrm{\beta } =4\sqrt{2}\mathrm{x} \\ \mathrm{\alpha }+\mathrm{\beta }=(2{\mathrm{x}}^4-1) \\ {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=66 \\ ={(\mathrm{\alpha }+\mathrm{\beta })}^2-\mathrm{\alpha }\times \mathrm{\beta }=66 \\ =32{\mathrm{x}}^{2}2(2{\mathrm{x}}^4-1)=66 \\ =32{\mathrm{x}}^2-4{\mathrm{x}}^4+2=66 \\ =16{\mathrm{x}}^2 -2{\mathrm{x}}^4+!=33 \\ =2{\mathrm{x}}^4-16{\mathrm{x}}^2+32=0 \\ ={\mathrm{x}}^4-8{\mathrm{x}}^2+16=0 \\ ={\mathrm{x}}^4-4{\mathrm{x}}^2-4{\mathrm{x}}^2+16=0 \\ = {\mathrm{x}}^2({\mathrm{x}}^{2^{ }}-4)-4({\mathrm{x}}^2-4)=0 \\ {(\mathrm{x}}^{} \end{gathered}$$

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