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 \(\boxed{x^2+4x^5+6x^8+4x^{11}+x^{14}}\). :)