alternate forms of $t^3 -1$

Question

I'm looking to understand how $t^3-1$ factorises to $(t-1)(t^2+t+1)$.

I know how to find the first factor $(t-1)$, but have trouble finding the second factor $(t^2+t+1)$.

I've tried doing long division of the original equation with the first factor, but that didn't work out.

Thanks for your help

Answer 1

$(t^3-1):(t-1)=t^2+t+1$

$-(t^3-t^2)$

$ t^2-1$

$ -(t^2-t)$

$ t-1$

$ -(t-1)$

$ 0$

The method proceeds as follows :

Divide the terms with the largest powers. Multiply the result with the denominator (here $t-1$) and subtract the result from the numerator (Here $t^3-1$).

Repeat this process with the remaining polynomial until the degree of the remaining polynomial after the subtraction is smaller than the degree of the denominator.

The degree of the remaining polynomial strictly decreases, so the process will terminate after a finite number of steps.

The last polynomial is the remainder of the division ($0$ in our case).

Answer 2

$$ t^3 - 1 = t^2 \cdot (t-1) + (t^2 -1) $$ $$ t^2 - 1 = t \cdot (t-1) + (t - 1)$$ $$ t -1 = 1 \cdot (t-1) + 0$$

This is the usual division algorithm so: $t^3 -1 = (t-1) \cdot (t^2 + t + 1)$

Answer 3

Suppose you already know that $t-1$ is a factor. A quick way to work out the other factor goes like this:

Since $t^3-1$ has degree $3$ and $t-1$ has degree $1$ then the other factor must have degree $3-1=2$.

So we expect $t^3-1=(t-1)(at^2+bt+c)$ for some $a,b,c$. The only term on the right hand side that contributes a cubic term is $t \times at^2 = at^3$ so comparing the coefficients gives $a=1$. Similarly we compare the constant term to get $c=1$. Now we have $t^3-1=(t-1)(t^2+bt+1)$. The $t^2$ term on the right has coefficient $-1+b$ but this must be $0$ from the left hand side. So $b=1$.

Thus $t^3-1=(t-1)(t^2+t+1)$.