The product sqrt[3]{2} cdot sqrt[4]{2} cdot s | vedclass

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(C) We have,$\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot (2^5)^{\frac{1}{12}}$$= 2^{\frac{1}{3}} \

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) We have,$\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot (2^5)^{\frac{1}{12}}$$= 2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot 2^{\frac{5}{12}}$Using the law of exponents $a^m \cdot a^n = a^{m+n}$,we get:$= 2^{\frac{1}{3} + \frac{1}{4} + \frac{5}{12}}$$= 2^{\frac{4+3+5}{12}}$$= 2^{\frac{12}{12}}$$= 2^1 = 2$Hence,$(c)$ is the correct answer.

The product $\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32}$ equals

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