sqrt[3]{-2197} times sqrt[3]{-125} + sqrt[3]{ | vedclass

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Question

(B) Step $1$: Calculate the cube root of $-2197$. Since $(-13)^3 = -2197$,we have $\sqrt[3]{-2197} = -13$.
Step $2$: Calculate the cube root of $-125

Vedclass · Accepted Answer

$\frac{523}{8}$.

Vedclass · Answer

(B) Step $1$: Calculate the cube root of $-2197$. Since $(-13)^3 = -2197$,we have $\sqrt[3]{-2197} = -13$.
Step $2$: Calculate the cube root of $-125$. Since $(-5)^3 = -125$,we have $\sqrt[3]{-125} = -5$.
Step $3$: Calculate the cube root of $\frac{27}{512}$. Since $3^3 = 27$ and $8^3 = 512$,we have $\sqrt[3]{\frac{27}{512}} = \frac{3}{8}$.
Step $4$: Substitute these values into the expression: $(-13) 	imes (-5) + \frac{3}{8}$.
Step $5$: Perform the multiplication: $65 + \frac{3}{8}$.
Step $6$: Simplify the expression: $\frac{65 	imes 8 + 3}{8} = \frac{520 + 3}{8} = \frac{523}{8}$.
Note: Upon reviewing the options,it appears there is a discrepancy. The calculated result is $\frac{523}{8}$. Given the provided options,none match exactly. However,if the expression was intended to be interpreted differently or if there is a typo in the options,the mathematical steps above are correct.

$\sqrt[3]{-2197} \times \sqrt[3]{-125} + \sqrt[3]{\frac{27}{512}}$

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