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Question
Find the cube of \[\frac{7}{9}\] .
Sum
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Solution
\[\because\] \[\left( \frac{m}{n} \right)^3 = \frac{m^3}{n^3}\]
\[\therefore\] \[\left( \frac{7}{9} \right)^3 = \frac{7^3}{9^3} = \frac{7 \times 7 \times 7}{9 \times 9 \times 9} = \frac{343}{729}\]
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