Question

When x=8, what is the value of `((x^-2)(27x^0))^(1/3)`?

Answer 1

`3/4`

Explanation:

`color(blue)(((x^-2)(27x^0))^(1/3)` `"when"` `color(brown)(x=8`

To, solve this, we need to know some exponential rules

`color(brown)(rArrx^0=1`

`color(brown)(rArrx^-z=1/(x^z)`

`color(brown)(rArrx^(1/y)=root(y)(x)`

Now, insert `8` into the problem

`rarr((8^-2)(27*8^0))^(1/3)`

Now, apply `color(brown)(x^0=1`

`rarrrarr((8^-2)(27))^(1/3)`

Apply `color(brown)(x^-z=1/x^z`

`rarr((1/8^2)(27))^(1/3)`

`rarr((1/64)(27))^(1/3)`

`rarr(27/64)^(1/3)`

Apply `color(brown)(x^(1/y)=root(y)(x)`

`rarrroot(3)(27/64)`

`rarrroot(3)((3xx3xx3)/(4xx4xx4))`

`color(green)(rArr3/4`

Hope that helps!!! ☻

Answer 2

`3/4`

Explanation:

Given: `(x^-2*27x^0)^(1/3)`

When `x=8`, we get:

`=(8^-2*27*8^0)^(1/3)`

`=(1/8^2*27*1)^(1/3)`

`=(1/64*27)^(1/3)`

`=(27/64)^(1/3)`

`=(27^(1/3))/(64^(1/3))`

`=(root(3)27)/(root(3)64)`

`=3/4`