$$x^2 -\sqrt 7 x +1 = 0 $$
$$\Rightarrow x^2 + 1 = \sqrt 7 x $$
$$\Rightarrow \dfrac {x^2 +1}{x} = \sqrt 7 $$
$$\Rightarrow x + \dfrac{1}{x} = \sqrt 7 $$Â
so $$ x^3 + x^-3Â = x^3 + \dfrac{1}{x^3}$$
$$\Rightarrow (x + \dfrac{1}{x})^3 - 3x\dfrac{1}{x}(x+\dfrac{1}{x})$$
$$\Rightarrow (\sqrt7)^3 - 3 \sqrt7 $$
$$\Rightarrow 7\sqrt 7 - 2\sqrt 7 $$
$$\Rightarrow 4 \sqrt7 $$ Ans
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