Answer:
x^3+3x^2+9x+27
Step-by-step explanation:
[tex]\dfrac{x^4-81}{x-3}=\dfrac{(x^2-9)(x^2+9)}{x-3}=\dfrac{(x-3)(x+3)(x^2+9)}{(x-3)}\\\\=(x+3)(x^2+9)=x^3+3x^2+9x+27[/tex]
_____
The factoring of the difference of squares is used (twice):
a² -b² = (a -b)(a +b)
