Prove the following


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

(a) $\sqrt{2}$

(b) 2

(c) $\sqrt[12]{2}$

(d) $\sqrt[12]{32}$

Thinking Process

Take the LCM of indices of three irrational numbers. Then, convert all individually in the form whose index will be equal to LCM.



LCM of 3,4 and $12=12$

$\sqrt[3]{2}=\sqrt[12]{2^{4}}$     $\left[\because \sqrt[m]{a}=\sqrt[m n]{a^{n}}\right]$


and         $\sqrt[12]{32}=\sqrt[12]{2^{5}}$

$\therefore$ Product of $\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32}=\sqrt[12]{2^{4}} \cdot \sqrt[12]{2^{3}} \cdot \sqrt[12]{2^{5}}=\sqrt[12]{2^{4} \cdot 2^{3} \cdot 2^{5}}$

$=12 \sqrt{2^{4+3+5}}=\sqrt[12]{2^{12}}=2^{12 \times \frac{1}{12}}=2 \quad\left[\because \sqrt[m]{a^{n}}=a^{n / m}\right]$

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