Binomial expansion of $\frac{1}{1+x+x^2}$ up to the first three terms

Question

Binomial expansion of $\frac{1}{1+x+x^2}$ up to the first three terms I am unsure where to start with this as i cannot put it into partial fractions, so don't really have an idea on where to start and pointers would be helpful.

Answer 1

$$\frac{1}{1+x+x^2}=\frac{1-x}{1-x^3} =(1-x)(1+x^3+x^6+...)$$

$$= 1-x+x^3 -x^4+...$$

Answer 2

As an alternative to the $\frac{1-x}{1-x^3}$ approach, you may know $\frac{1}{1-y}=1+y+y^2+y^3+y^4+\cdots$

so letting $y=-x-x^2$ you get $$\frac{1}{1+x+x^3} = 1 +(-x-x^2) +(-x-x^2)^2 + (-x-x^2)^3 + (-x-x^2)^4 + \cdots \\= 1 - x -x^2 + x^2 +2x^3 +x^4 - x^3 - 3x^4 - \cdots +x^4+\cdots \\ =1 - x +x^3 -x^4 +\cdots$$