How to decompose $x^3-1$ I need to decompose $x^3-1$, I know the Binomial theorem, and finding roots of a polynomial, how should I approach this? 
 A: Hint: what's the 'obvious' root of your polynomial? Do you know the remainder theorem?
A: As $(a-b)^3=a^3-b^3-3ab(a-b)$
$a^3-b^3=\cdots=(a-b)(a^2+ab+b^2)$
More generally. $$a^{n+1}-b^{n+1}=(a-b)(a^n+a^{n-1}b+\cdots+ab^{n-1}+b^n)$$ where integer $n\ge1$
A: Let $$f(x)=x^3 - 1$$
$f(1)=0$ is a solution.
Hence $(x-1)$ is a factor of $f(x)$
You can then either factorise $f(x)$ by hand, or find the other factor via long division of $\frac{f(x)}{x-1}$
A: You have $(x-1)(x^2+x+1)$ now we need a root for the second polynomial.
We have $x^2+x+1=0\iff x^2+x+\frac{1}{4}=\frac{-3}{4}\iff(x+\frac{1}{2})^2=\frac{-3}{4}\iff x+\frac{1}{2}=\frac{\pm\sqrt{-3}}{2}$
$\iff x=\frac{\pm\sqrt{-3}-1}{2}$
So the polynomial is equal to:
$(x-1)(x+\frac{1+\sqrt{3}i}{2})(x+\frac{1-\sqrt{3}i}{2})$
A: We can just use binomial theorem:
Observe
$$(x-1)^3=x^3-3x^2+3x-1$$
So $$\begin{align}x^3-1&=(x-1)^3+3x^2-3x\\
&=(x-1)^3+3x(x-1)\\
&=(x-1)((x-1)^2+3x)\\
&=(x-1)(x^2+x+1)\end{align}$$
Hope this helps.
P.S.
If factoring $x^2+x+1$ is needed, recall we have a formula for quadratic polynomials.
