Question

If X^-2 =64 then what will be the value of 1/x^3 + x^0?

Answer 1

`513, or -511`.

Explanation:

`x^-2=64 rArr 1/x^2=64 rArr x^2=1/64 rArr x=+-1/8`.

`rArr 1/x=+-8`.

`:. 1/x^3+x^0=(1/x)^3+1=(+-8)^3+1=+-512+1`.

`:." The Reqd. Value="513, or -511`.

Answer 2

Please see another method in resolving below;

Explanation:

Given;

`x^-2 = 64`

Using Indices

Recall: `x^-1 = 1/x`

`x^-2 = 64`

`1/x^2 = 64`

`1/x^2 = 64/1`

Cross multiplying..

`64 xx x^2 = 1 xx 1`

`64 cdot x^2 = 1`

Divide both sides by `64`

`(64 cdot x^2)/64 = 1/64`

`(cancel64 cdot x^2)/cancel64 = 1/64`

`x^2 = 1/64`

Squareroot both sides..

`sqrtx^2 = sqrt(1/64)`

`x = 1/8`

Hence;

`1/x^3 + x^0`

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

Recall: `x^0 = 1`

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

`1/(1/512) + 1`

`1 div 1/512 + 1`

`1 xx 512/1 + 1 color(white)(xxxxx)color(green)(512/1) -> "Transposed"`

`1 xx 512 + 1`

`512 + 1`

`513 -> "Answer"`

Hope this helps and its well understood!