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If \(x+\frac{1}{x}=7\), what is the value of \(x^3+\frac{1}{x^3}\)?

Answer and explanation

Correct answer: (322)

The direct answer is option B, 322. Let \\(a=x\\) and \\(b=\\frac{1}{x}\\). We are given \\(a+b=7\\). Also, \\(ab=x\\cdot\\frac{1}{x}=1\\), provided \\(x\\neq0\\). Use the identity \\(a^3+b^3=(a+b)^3-3ab(a+b)\\). Substitution gives \\(7^3-3(1)(7)=343-21=322\\). Thus option B is correct. Option A, 336, does not subtract the required \\(3ab(a+b)\\) term. Option B follows the identity exactly. Option C, 343, is only \\(7^3\\); it forgets that \\(a^3+b^3\\) is not simply \\((a+b)^3\\). Option D, 364, results from an incorrect operation and is larger than the correct result. Do not try to find x itself; the identity gives the requested expression directly. Memory cue: if \\(x+1/x=7\\), then the cubes are \\(7^3-3\\times7\\) because the product is 1.

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