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

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Answer 1 · 2018-03-14T22:28:35

Ok

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

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 · 2018-03-14T22:35:30

See the 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: $LHS=RHS$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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 $x in RR and x!=-2$

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