Answer 1

We have to find the value of `x^(1/3+ x^0)` if  `x^-2 = 64`

Consider,

 `x^-2 = 2^6`

 `1/x^2 = 2^6`

Multiply  `1/2`on both sides of powers we get 

`1/x^(2 xx 1/2) = 2^(6x1/2)`

`1/x^(2 xx 1/2) = 2^(6x1/2)`

`1/x = 2^3`

`1/x = 8/ 1`

By taking reciprocal on both sides we get,

 `1/8 = x `

Substituting `1/8` in `x^(1/3 + x^0)`we get

`= (1/8)^(1/3) + (1/8)^0`

`= (1/(2^3))^(1/3) + (1/8)^0`

`= 1/(2^(3xx 1/3)) +1`

` = 1/2^1 + 1` 

`=1/2 +1`

By taking least common multiply we get 

`= 1/2 + (1 xx 2)/(1 xx 2)`

` = 1/2 + 2/2 `

`= (1+2)/2`

`= 3/2`