We just started proofs in my mathematical structures class so this is the first time i am dealing with proofs.

The question i am having trouble proving is:

**Suppose x is a real number and x $\displaystyle \neq $ 0. Prove that if $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} $ then $\displaystyle x \neq 8. $**
This is what i have started:

**Givens:**
x is a real number

x $\displaystyle \neq $ 0

$\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} $

**Goals:**
$\displaystyle x \neq 8. $

i feel like to complete this proof you would solve the equation $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} $ for x and then this would show that whatever this value of x is, it will not be equal to 8.

The problem i am running into is that i cannot solve the equation $\displaystyle \frac{x^{1/3} + 5}{x^2+6} = \frac{1}{x} $ for x equaling a value, so i feel like i may be going about this wrong.

Any help / tips for proofs would be very great

Thank you