If \(x + \frac{1}{x} = 3\), find the value of \(x^{3} + \frac{1}{x^{3}}\).
18
To find the value of \(x^{3} + \frac{1}{x^{3}}\) given that \(x + \frac{1}{x} = 3\), we can use algebraic identities and manipulation.
- First, recall the identity for the cube of a binomial:
\[\left(x + \frac{1}{x}\right)^{3} = x^{3} + \frac{1}{x^{3}} + 3\left(x + \frac{1}{x}\right)\]
- Substitute \(x + \frac{1}{x} = 3\) into the identity:
\[3^{3} = x^{3} + \frac{1}{x^{3}} + 3 \cdot 3\]
- Calculate the left side of the equation:
\[27 = x^{3} + \frac{1}{x^{3}} + 9\]
- Isolate \(x^{3} + \frac{1}{x^{3}}\) by subtracting 9 from both sides:
\[x^{3} + \frac{1}{x^{3}} = 27 - 9 = 18\]
Therefore, the value of \(x^{3} + \frac{1}{x^{3}}\) is 18, which matches the correct answer option.
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