Discussion
If \((\frac{3}{8})^{-5} \times (\frac{16}{21})^{-5} = (\frac{2}{7})^{x}\), then the value of x3 is
Answer: D
The value of \(x^3\) is −125.
Both factors have the same power, so they can be multiplied inside one bracket:
\[ \begin{aligned} \left(\frac{3}{8} \times \frac{16}{21}\right)^{-5} &= \left(\frac{2}{7}\right)^{x} \\ \frac{3 \times 16}{8 \times 21} &= \frac{48}{168} = \frac{2}{7} \\ \left(\frac{2}{7}\right)^{-5} &= \left(\frac{2}{7}\right)^{x} \end{aligned} \]
So \(x = -5\) and \(x^3 = (-5)^3 = -125\). The option 125 forgets that the cube of a negative number is negative, and ±27 would need \(x = \pm 3\).
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