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If \(f\left(x+\frac{1}{x}\right)=x^3+\frac{1}{x^3}\) then \(f(\sqrt{3})\) is equal to |
\(0\) \(1\) \(\sqrt{3}\) \(3\sqrt{3}\) |
\(0\) |
$f\left(x+\frac{1}{x}\right)=\frac{1}{x^3}+x^3$ $=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)$ so $f(y)=y^3-3y$ at $y=\sqrt{3}$ $f(\sqrt{3})={\sqrt{3}}^2-3\sqrt{3}=0$ |
