Reduce to lowest terms:
$\displaystyle \frac {x^3+x^2+x+1}{x^3+3x^3+3x+1}$
I don't see how this can be factored to be reduced. Thanks.
The answer is $\displaystyle \frac{x^2+1}{x^2+2x+1}$
For the numerator:
$\displaystyle f(x)=x^3+x^2+x+1$
You could use the rational roots theorem, which states that if the given polynomial has a rational root, it will come from the list $\displaystyle x=\pm1$. We can see that $\displaystyle x=+1$ cannot work, hence we find:
$\displaystyle f(-1)=-1+1-1+1=0$
So, we know $\displaystyle x+1$ is a factor of $\displaystyle f(x)$. Use of division finds:
$\displaystyle f(x)=(x+1)(x^2+1)$
Now, for the denominator, we should recognize the binomial coefficients arising from the cube of a binomial, i.e.:
$\displaystyle x^3+3x^2+3x+1=(x+1)^3$ and so we may state:
$\displaystyle \frac{x^3+x^2+x+1}{x^3+3x^2+3x+1}=\frac{(x+1)(x^2+ 1)}{(x+1)^3}=\frac{x^2+1}{(x+1)^2}=\frac{x^2+1}{x^ 2+2x+1}$