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Cube Root Using Prime Factorisation

Is 243 a perf...

Question

Is 243 a perfect cube

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Solution

Dear Student,

$243 = 3 \times 3 \times 3 \times 3 \times 3 n o w f i n d i n g c u b e r o o t o f 243 = \sqrt[3]{243} = \sqrt[3]{3 \times 3 \times 3 \times 3 \times 3} = \sqrt[3]{3^{3} \times 3^{2}} = 3 \times \sqrt[3]{3^{2}} = 3 \times \sqrt[3]{9} H e n c e 243 i s n o t a p e r f e c t c u b e$

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