Simplify: frac{8^{frac{1}{3}} times 16^{frac{ | vedclass

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Question

(B) To simplify the expression $\frac{8^{\frac{1}{3}} 	imes 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$,we express each base as a power of $2$:$8 = 2^3$,$1

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) To simplify the expression $\frac{8^{\frac{1}{3}} 	imes 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$,we express each base as a power of $2$:$8 = 2^3$,$16 = 2^4$,and $32 = 2^5$.Substituting these into the expression:$\frac{(2^3)^{\frac{1}{3}} 	imes (2^4)^{\frac{1}{3}}}{(2^5)^{-\frac{1}{3}}}$Using the power rule $(a^m)^n = a^{m 	imes n}$:$\frac{2^{3 	imes \frac{1}{3}} 	imes 2^{4 	imes \frac{1}{3}}}{2^{5 	imes -\frac{1}{3}}} = \frac{2^1 	imes 2^{\frac{4}{3}}}{2^{-\frac{5}{3}}}$Using the product rule $a^m 	imes a^n = a^{m+n}$ in the numerator:$\frac{2^{1 + \frac{4}{3}}}{2^{-\frac{5}{3}}} = \frac{2^{\frac{7}{3}}}{2^{-\frac{5}{3}}}$Using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$:$2^{\frac{7}{3} - (-\frac{5}{3})} = 2^{\frac{7}{3} + \frac{5}{3}} = 2^{\frac{12}{3}} = 2^4 = 16$.

Simplify: $\frac{8^{\frac{1}{3}} \times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$

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