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Question
Find the cube of \[\frac{12}{7}\] .
Sum
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Solution
\[\because\] \[\left( \frac{m}{n} \right)^3 = \frac{m^3}{n^3}\]
\[\therefore\] \[\left( \frac{12}{7} \right)^3 = \frac{{12}^3}{7^3} = \frac{12 \times 12 \times 12}{7 \times 7 \times 7} = \frac{1728}{343}\]
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