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

2 Answers
Mar 14, 2018

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

Mar 14, 2018

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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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