If \(x + \dfrac{2}{x} = 7\), find the value of \(x^3 + \dfrac{8}{x^3}\).
301
Use the cube identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\).
With \(a = x\) and \(b = \dfrac{2}{x}\):
\(\left(x + \dfrac{2}{x}\right)^3 = x^3 + \dfrac{8}{x^3} + 3 \cdot x \cdot \dfrac{2}{x}\left(x + \dfrac{2}{x}\right)\)
\(7^3 = x^3 + \dfrac{8}{x^3} + 6 \times 7\)
\(343 = x^3 + \dfrac{8}{x^3} + 42\)
\(x^3 + \dfrac{8}{x^3} = 343 - 42 = 301\)
Hence the answer is 301.
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