Algebra

Let's put them all of them with denominator x. 

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

Factoring the cube first, we get

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

Canceling out the denominator, we get

$\left(x^3+1\right)x^2\left(x+1\right)^3\left(x^2-x+1\right)^3$

Expanding everything, we finally get

$x^{14}+4x^{11}+6x^8+4x^5+x^2$

That's our final answer!

Thanks! :)

$(x^2 + 1/x)*(x + x^4)^3$

i then reformatted stuff to get$(\frac{x^3+1}{x})(x^3+3x^6+3x^9+x^{12})$

i further expanded to get$\frac{x^3+4x^9+6x^9+4x^{12}+x^{15}}{x}$

then just substract one from all the powers in the numerator to get the final answer of $\boxed{x^2+4x^5+6x^8+4x^{11}+x^{14}}$. :)