All the marked sums please . Tommorow is my maths half yearly examination

Dear student
Ans8.
$$\begin{gathered} We have, \\ {a}^2+\frac{1}{a^2}=4 \\ Then, \\ 2a^3+\frac{2}{a^3} \\ =2\left(a^3+\frac{1}{a^3}\right) \\ =2\left[\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-\left(a\right)\left(\frac{1}{a}\right)\right)\right] \left[{\left(a+b\right)}^3=\left(a+b\right)(a^2+b^2-ab\right] \\ =2\left[\left(\sqrt{a^2+\frac{1}{a^2}+2\left(a\right)\left(\frac{1}{a}\right)}\right)\left(a^2+\frac{1}{a^2}-1\right)\right] \left[{\left(a+b\right)}^2=\left(a^2+b^2+2ab\right)\right] \\ =2\left[\left(\sqrt{4+2}\right)\left(4-1\right)\right] \\ =2\left(\sqrt{6}\right)\left(3\right) \\ =6\sqrt{6} \end{gathered}$$
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