Find a compact form for generating function of the sequence 1, 8, 27,.........., k^3,.........

The generating function is [tex]f(x)[/tex] where[tex]f(x)=\displaystyle\sum_{k=0}^\infty a_kx^k[/tex]with [tex]a_k=k^3[/tex] for [tex]k\ge0[/tex].Recall that for [tex]|x|<1[/tex], we have[tex]g(x)=\dfrac1{1-x}=\displaystyle\sum_{k=0}^\infty x^k[/tex]Taking the derivative gives[tex]g'(x)=\dfrac1{(1-x)^2}=\displaystyle\sum_{k=1}^\infty kx^{k-1}=\sum_{k=0}^\infty(k+1)x^k[/tex][tex]\implies g'(x)-g(x)=\dfrac x{(1-x)^2}=\displaystyle\sum_{k=0}^\infty kx^k[/tex]Taking the derivative again, we get[tex]g''(x)=\dfrac2{(1-x)^3}=\displaystyle\sum_{k=2}^\infty k(k-1)x^{k-2}=\sum_{k=0}^\infty(k^2+3k+2)x^k[/tex][tex]\implies g''(x)-3g'(x)+g(x)=\dfrac{x^2+x}{(1-x)^3}=\displaystyle\sum_{k=0}^\infty k^2x^k[/tex]Take the derivative one last time to get[tex]g'''(x)=\dfrac6{(1-x)^4}=\displaystyle\sum_{k=3}^\infty k(k-1)(k-2)x^{k-3}=\sum_{k=0}^\infty(k^3+6k^2+11k+6)x^k[/tex][tex]\implies g'''(x)-6g''(x)+7g'(x)-g(x)=\dfrac{x^3+4x^2+x}{(1-x)^4}=\displaystyle\sum_{k=0}^\infty k^3x^k[/tex]So the generating function is[tex]\boxed{f(x)=\dfrac{x^3+4x^2+x}{(1-x)^4}}[/tex]