Question
If \(x^{3} = \frac{9^{3} \times 6^{3}}{8}\), \(x \gt 0\), then what is the value of x ?
This question was previously asked in
UPSSSC PET 2026 Question Paper (25-Oct-2026) (Shift-2)
The correct answer is
27
Take the cube root of both sides; since x is positive, only the real positive cube root is needed, and the cube root of a product or quotient is the product or quotient of the cube roots.
\(x = \sqrt[3]{\frac{9^{3} \times 6^{3}}{8}} = \frac{9 \times 6}{2}\)
\(x = \frac{54}{2} = 27\)
Check: \(27^{3} = 19683\) and \(\frac{729 \times 216}{8} = \frac{157464}{8} = 19683\), so both sides agree.
Hence, 27.
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