({x^5})^{1/3} times (16{x^3})^{2/3} times lef | vedclass

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(D) Given expression: $({x^5})^{1/3} 	imes (16{x^3})^{2/3} 	imes \left( \frac{1}{4}{x^{4/9}} ight)^{-3/2}$ Step $1$: Simplify each term using expo

Vedclass · Accepted Answer

None of these

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(D) Given expression: $({x^5})^{1/3} 	imes (16{x^3})^{2/3} 	imes \left( \frac{1}{4}{x^{4/9}} ight)^{-3/2}$ Step $1$: Simplify each term using exponent rules $(a^m)^n = a^{mn}$. $({x^5})^{1/3} = x^{5/3}$ $(16{x^3})^{2/3} = (2^4)^{2/3} 	imes (x^3)^{2/3} = 2^{8/3} 	imes x^2$ $\left( \frac{1}{4}{x^{4/9}} ight)^{-3/2} = (4^{-1} 	imes x^{4/9})^{-3/2} = (2^{-2})^{-3/2} 	imes (x^{4/9})^{-3/2} = 2^3 	imes x^{-6/9} = 8 	imes x^{-2/3}$ Step $2$: Multiply the simplified terms together. $x^{5/3} 	imes 2^{8/3} 	imes x^2 	imes 2^3 	imes x^{-2/3}$ $= 2^{8/3 + 3} 	imes x^{5/3 + 2 - 2/3}$ $= 2^{17/3} 	imes x^{3/3 + 2} = 2^{17/3} 	imes x^3$ Since $2^{17/3} 
eq 8$,the result is $2^{17/3}x^3$. Thus,the correct option is $D$.

$({x^5})^{1/3} \times (16{x^3})^{2/3} \times \left( \frac{1}{4}{x^{4/9}} \right)^{-3/2} = ?$

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