Question

Express the following rational expression to its lowest form.
$\frac{x^3-27}{x^2-9}$

• A. $\frac{(x^2-9x+9)}{(x+3)}$
• B. $\frac{(x^2+3x+9)}{(x+3)}$
• C. $\frac{(x^2+3x+6)}{(2x+3)}$
• D. None of the above

Answer 1

The correct option is B $\frac{(x^2+3x+9)}{(x+3)}$

$\frac{x^3-27}{x^2-9}$

Here $p\left(x\right)=x^3-27=x^3-3^3$
= $(x-3)(x^2+3x+9)$

$q\left(x\right)=x^2-9=x^2-3^2$
= $(x+3)(x-3)$

Now $\frac{p\left(x\right)}{q\left(x\right)}=\frac{(x-3)(x^2+3x+9)}{(x+3)(x-3)}$

Cancelling $(x-3)$ from numerator and denominator,

$\frac{p\left(x\right)}{q\left(x\right)}=\frac{(x^2+3x+9)}{(x+3)}$