Question

If x^-2=64 then find x^1/3+x^0

Answer 1

Step-by-step explanation:

${x}^{ - 2} = 64$

$x = { ({8}^{2}) }^{ \frac{ - 1}{2} }$

$x = {8}^{ - 1} = \frac{1}{8}$

SUBSTITUTE X IN EQUATION

$( { \frac{1}{8}) }^{ \frac{1}{3} } + ({ \frac{1}{8}) }^{0}$

$( { {2}^{ - 3} )}^{ \frac{1}{3} } + 1$

${2}^{ - 1} + 1$

$\frac{1}{2} + 1$

$\frac{3}{2}$

Answer 2

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