Simplify: 64^{-frac{1}{3}} + 64^{frac{1}{3}} | vedclass

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Question

(A) Given expression: $64^{-\frac{1}{3}} + 64^{\frac{1}{3}} - 64^{\frac{2}{3}}$
First,express $64$ as a power of $4$: $64 = 4^3$.
Substitute $4^3$ i

Vedclass · Accepted Answer

$-\frac{47}{4}$

Vedclass · Answer

(A) Given expression: $64^{-\frac{1}{3}} + 64^{\frac{1}{3}} - 64^{\frac{2}{3}}$
First,express $64$ as a power of $4$: $64 = 4^3$.
Substitute $4^3$ into the expression:
$(4^3)^{-\frac{1}{3}} + (4^3)^{\frac{1}{3}} - (4^3)^{\frac{2}{3}}$
Using the power rule $(a^m)^n = a^{m \cdot n}$:
$4^{3 \cdot (-\frac{1}{3})} + 4^{3 \cdot \frac{1}{3}} - 4^{3 \cdot \frac{2}{3}}$
Simplify the exponents:
$4^{-1} + 4^1 - 4^2$
Calculate the values:
$\frac{1}{4} + 4 - 16$
Combine the terms:
$\frac{1}{4} - 12$
Convert to a common denominator:
$\frac{1 - 48}{4} = -\frac{47}{4}$
Wait,re-evaluating the calculation: $4 - 16 = -12$. $\frac{1}{4} - 12 = \frac{1 - 48}{4} = -\frac{47}{4}$.
Re-checking the options provided: The calculated result is $-\frac{47}{4}$. Since this is not in the options,let's re-read the question. If the expression was $64^{-\frac{1}{3}} (64^{\frac{1}{3}} - 64^{\frac{2}{3}})$,the result would differ. Given the standard form,the result is $-\frac{47}{4}$. Assuming a typo in the question options,we will provide the correct mathematical derivation.

Simplify: $64^{-\frac{1}{3}} + 64^{\frac{1}{3}} - 64^{\frac{2}{3}}$

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