Answer:
[tex](\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)[/tex]
Step-by-step explanation:
We have been given that
[tex]f(x)=x^3-27,g(x)=3x-9[/tex]
We can use the formula for difference of cubes to simplify the function f(x)
difference of cubes - [tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
[tex]f(x)=x^3-27\\\\=x^3-3^3\\\\=(x-3)(x^2+3x+9)[/tex]
And g(x) can be written as
[tex]g(x)=3x-9\\=3(x-3)[/tex]
Thus, we have
[tex](\frac{f}{g})(x)=\frac{(x-3)(x^2+3x+9)}{3(x-3}[/tex]
On cancelling the common factors, we get
[tex](\frac{f}{g})(x)=\frac{1}{3}(x^2+3x+9)[/tex]
