$The value of the expression {(\log_{21} 3)}^{3} + \log_{21} 27 \cdot \log_{21} 7 + {(\log_{21} 7)}^{3} is$

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Detailed Solution

$\begin{array}{l} Let \log_{21} 3 = a and \log_{21} 7 = b \\ a + b = 1 \\ \Rightarrow (a + b)^{3} = 1 \\ \Rightarrow a^{3} + b^{3} + 3 ab (a + b) = 1 \\ Hence, a^{3} + 3 ab + b^{3} = 1 \end{array}$

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