Show that each of the following numbers is a perfect cube. Also find the number whose cube is the given number:

1728

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5832

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13824

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35937

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Solution

The correct option is A $1728$
$\sqrt[3]{1728} = \sqrt[3]{2^{6} \times 3^{3}} = 12 \sqrt[3]{5832} = \sqrt[3]{3^{6} \times 2^{3}} = 18 \sqrt[3]{13824} = \sqrt[3]{2^{9} \times 3^{3}} = 24 \sqrt[3]{35937} = \sqrt[3]{3^{3} \times 11^{3}} = 33$

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Q. Find the given number is a perfect cube or not.

13824

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Show that each of the following numbers is a perfect cube. Also find the number whose cube is the given number:

The correct option is A 17283√1728=3√26×33=123√5832=3√36×23=183√13824=3√29×33=243√35937=3√33×113=33

The correct option is A $1728$
$\sqrt[3]{1728} = \sqrt[3]{2^{6} \times 3^{3}} = 12 \sqrt[3]{5832} = \sqrt[3]{3^{6} \times 2^{3}} = 18 \sqrt[3]{13824} = \sqrt[3]{2^{9} \times 3^{3}} = 24 \sqrt[3]{35937} = \sqrt[3]{3^{3} \times 11^{3}} = 33$