Question

How do you prove that `(x^3+9)/(x^3+8) = 1+1/(x^3+8)` where `x` is not equal to `-2`?

Answer 1

Ok

Explanation:

`(x^3+9)/(x^3+8)=1+1/(x^3+8)`

You take away from both sides `1/(x^3+8)`

`(x^3+9)/(x^3+8)-1/(x^3+8)=1`

Since the denominator is the same you can sum the numerators

`(x^3+9-1)/(x^3+8)=1`

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

Which is of course true as you can see

Answer 2

See the explanation

Explanation:

Given that the equation becomes undefined at `x=-2` because `x^3+8=0`. You are 'not allowed' to divide by 0.

Stating the obvious:

Suppose `(x^3+9)/(x^3+8)=1+1/(x^3+8)`

Then as long as we follow the rules of mathematics we will always have: `LHS=RHS`
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Note that `1= (x^3+8)/(x^3+8)` then by substituting for 1 we have:

`(x^3+9)/(x^3+8)=(x^3+8)/(x^3+8)+1/(x^3+8)`

Multiply all of both sides by `(x^3+8)`

`x^3+9=x^3+8+1`

`x^3+9=x^3+9`

Thus: `LHS=RHS` for `x in RR and x!=-2`