Algebraically prove that X^3+10/x^3+9=1+ 1/x^3+9 where x is not equal to -3
[tex] \dfrac{x^3+10}{x^3+9}=1+\dfrac{1}{x^3+9} \\ \\ \\ \dfrac{x^3+10}{x^3+9}=\dfrac{(x^3+9)+1}{x^3+9}\\ \\ \\ \dfrac{x^3+10}{x^3+9}=\dfrac{x^3+10}{x^3+9} \\ \\ \\x\in R \qquad x\in (-\infty, +\infty) [/tex]
