Getting $(\frac{1-x^{6}}{1-x})^3$ from $(1 + x + x^2 + x^3+ x^4+ x^5)^3$ using generating functions

Question

I came across when reading my solution that the expression $(1 + x + x^2 + x^3+ x^4+ x^5)^3$ simplifies to $(\frac{1-x^{6}}{1-x})^3$ using generating function. I'm not sure how they got this.

I know the following series.

$$\frac{1}{1-x}=(1+x+x^2 + x^3 + x^4 + ...)$$

$$\frac{1}{1-x^6}=(1+x^6+x^{12} + x^{18} + x^{24} + ...)$$

EDIT: Mistake in question.

Answer 1

No generating functions involved. Just multiply out

$$(1-x)(1+x+\cdots+x^5) = (1+x+\cdots+x^5)-x(1+x+\cdots+x^5)=1-x^6$$

noticing that you get a lot of cancelation of terms. Make sure you see this. It also works for any number in the place of 6, provided the sequence of $x$ terms go up to $x^{n-1}$.

Answer 2

Note that $$\dfrac{1-x^6}{1-x} = \dfrac{(1-x)(x^5+x^4+x^3+x^2+x+1)}{1-x}$$

Answer 3

Hint: It is just the nth sum of a geometric series:

$$\sum_{k=0}^{n-1} x^k=\frac{1-x^n}{1-x}$$

Answer 4

$$1-x^6=1^3-(x^2)^3=(1-x^2)(1+x^2+x^4)=$$ $$=(1-x)(1+x)(1+x^2+x^4)=$$ $$=(1-x)(1+x^2+x^4+x+x^3+x^5)=$$ $$=(1-x)(1+x+x^2+x^3+x^4+x^5)$$