## Task B-1.1.

Simplify the fraction $\frac{(x+1)^{4}-4\left(x+x^{2}\right)^{2}+4 x^{2}-(1+x)^{2}}{3 x^{2}+x}$ if $x \neq 0, x \neq-\frac{1}{3}$.
## Solution.

Let's factorize the numerator and the denominator of the given fraction.

Factor out $x$ in the denominator and from the second parenthesis in the numerator:

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

Now, in the numerator, factor out the common factor from the first and last term, and from the second and third term, respectively:

$=\frac{(x+1)^{2}\left[(x+1)^{2}-1\right]-4 x^{2}\left[(1+x)^{2}-1\right]}{x(3 x+1)}$ 1 point

$=\frac{\left[(x+1)^{2}-1\right]\left[(x+1)^{2}-4 x^{2}\right]}{x(3 x+1)}$

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

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

Now we can simplify the fraction, so the solution is

$=(x+2)(1-x)$.