Find the value of frac{8^{1/3} times 16^{1/3} | vedclass

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(B) To find the value of $\frac{8^{1/3} 	imes 16^{1/3}}{32^{-1/3}}$,we express each base as a power of $2$:$8 = 2^3$,$16 = 2^4$,and $32 = 2^5$.Substi

Vedclass · Accepted Answer

$16$

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(B) To find the value of $\frac{8^{1/3} 	imes 16^{1/3}}{32^{-1/3}}$,we express each base as a power of $2$:$8 = 2^3$,$16 = 2^4$,and $32 = 2^5$.Substituting these into the expression:$\frac{(2^3)^{1/3} 	imes (2^4)^{1/3}}{(2^5)^{-1/3}} = \frac{2^{3 	imes 1/3} 	imes 2^{4 	imes 1/3}}{2^{5 	imes -1/3}} = \frac{2^1 	imes 2^{4/3}}{2^{-5/3}}$.Using the laws of exponents $a^m 	imes a^n = a^{m+n}$ and $\frac{a^m}{a^n} = a^{m-n}$:$= 2^{1 + 4/3 - (-5/3)} = 2^{1 + 4/3 + 5/3} = 2^{1 + 9/3} = 2^{1 + 3} = 2^4$.$2^4 = 16$.

Find the value of $\frac{8^{1/3} \times 16^{1/3}}{32^{-1/3}}$.

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