Find the cube root of each of the following numbers:
(i) $8 \times 125$
(ii) $-1728 \times 216$
(iii) $-27 \times 2744$
(iv) $-729 \times -15625$

To find: 

We have to find the cube roots of the given numbers.

Solution:

(i) $ \sqrt[3]{8 \times 125}=\sqrt[3]{2 \times 2 \times 2 \times 5 \times 5 \times 5}$

$=\sqrt[3]{2^{3} \times 5^{3}}$

$=2 \times 5$

$=10$

(ii) $\sqrt[3]{-1728 \times 216}=-\sqrt[3]{1728 \times 216}$

$=-\sqrt[3]{1728} \times \sqrt[3]{216}$

$=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3} \times \sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3}$

$=\sqrt[3]{2^{3} \times 2^{3} \times 3^{3}} \times \sqrt[3]{2^{3} \times 3^{3}}$

$=-(2 \times 2 \times 3) \times(2 \times 3)$

$=-12 \times 6$

$=-72$

(iii) $\sqrt[3]{-27 \times 2744}=-\sqrt[3]{27 \times 2744}$

$=-\sqrt[3]{27} \times \sqrt[3]{2744}$

$=-(\sqrt[3]{3 \times 3 \times 3}) \times(\sqrt[3]{2 \times 2 \times 2 \times 7 \times 7 \times 7})$

$=-(\sqrt[3]{3^{3}} \times \sqrt[3]{2^{3} \times 7^{3}})$

$=-(3 \times 2 \times 7)$

$=-42$

(iv) $\sqrt[3]{(-729) \times(-15625)}$

$=\sqrt[3]{729 \times 15625}$

$=\sqrt[3]{729} \times \sqrt[3]{15625}$

$=\sqrt[3]{3 \times 3 \times 3 \times 3 \times 3 \times 3} \times \sqrt[3]{5 \times 5 \times 5 \times 5 \times 5 \times 5}$

$=\sqrt[3]{3^{3} \times 3^{3}} \times \sqrt[3]{5^{3} \times 5^{3}}$

$=(3 \times 3) \times(5 \times 5)$

$=9 \times 25$

$=225$

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Updated on: 10-Oct-2022

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