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

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Answer 1 · 2018-03-14T22:28:35

Ok

$\frac{x^{3} + 9}{x^{3} + 8} = 1 + \frac{1}{x^{3} + 8}$ $(x^3+9)/(x^3+8)=1+1/(x^3+8)$

You take away from both sides $\frac{1}{x^{3} + 8}$ $1/(x^3+8)$

$\frac{x^{3} + 9}{x^{3} + 8} - \frac{1}{x^{3} + 8} = 1$ $(x^3+9)/(x^3+8)-1/(x^3+8)=1$

Since the denominator is the same you can sum the numerators

$\frac{x^{3} + 9 - 1}{x^{3} + 8} = 1$ $(x^3+9-1)/(x^3+8)=1$

$\frac{x^{3} + 8}{x^{3} + 8} = 1$ $(x^3+8)/(x^3+8)=1$

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 $x = - 2$ $x=-2$ because $x^{3} + 8 = 0$ $x^3+8=0$ . You are 'not allowed' to divide by 0.

Stating the obvious:

Suppose $\frac{x^{3} + 9}{x^{3} + 8} = 1 + \frac{1}{x^{3} + 8}$ $(x^3+9)/(x^3+8)=1+1/(x^3+8)$

Then as long as we follow the rules of mathematics we will always have: $L H S = R H S$ $LHS=RHS$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Note that $1 = \frac{x^{3} + 8}{x^{3} + 8}$ $1= (x^3+8)/(x^3+8)$ then by substituting for 1 we have:

$\frac{x^{3} + 9}{x^{3} + 8} = \frac{x^{3} + 8}{x^{3} + 8} + \frac{1}{x^{3} + 8}$ $(x^3+9)/(x^3+8)=(x^3+8)/(x^3+8)+1/(x^3+8)$

Multiply all of both sides by $(x^{3} + 8)$ $(x^3+8)$

$x^{3} + 9 = x^{3} + 8 + 1$ $x^3+9=x^3+8+1$

$x^{3} + 9 = x^{3} + 9$ $x^3+9=x^3+9$

Thus: $L H S = R H S$ $LHS=RHS$ for $x in RR and x!=-2$

How do you prove that (x^3+9)/(x^3+8) = 1+1/(x^3+8)x3+9x3+8=1+1x3+8(x^3+9)/(x^3+8) = 1+1/(x^3+8) where xxx is not equal to -2−2-2?