If \(x + \frac{1}{x} = 2\sqrt{3}\), then find the value of \(x^{3} + \frac{1}{x^{3}}\).
\(18\sqrt{3}\)
We are given \(x+\frac{1}{x}=2\sqrt{3}\) and asked for \(x^{3}+\frac{1}{x^{3}}\).
Use the cube identity \(a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)\). With \(a=x,\;b=\frac{1}{x}\), the product \(ab=x\cdot \frac{1}{x}=1\).
So the identity becomes \(x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)\).
Substitute \(x+\frac{1}{x}=2\sqrt{3}\): the expression is \(\left(2\sqrt{3}\right)^{3}-3\left(2\sqrt{3}\right)\).
Compute the cube: \(\left(2\sqrt{3}\right)^{3}=2^{3}\times(\sqrt{3})^{3}=8\times 3\sqrt{3}=24\sqrt{3}\).
Compute the second term: \(3\times 2\sqrt{3}=6\sqrt{3}\), then subtract: \(24\sqrt{3}-6\sqrt{3}=18\sqrt{3}\).
The key formula is the sum-of-cubes identity applied to \(x\) and its reciprocal. Hence \(x^{3}+\frac{1}{x^{3}}\) equals 18√3.
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