Sum the first n^3 how provide with using mathematical induction? I having next formula and need solving with helps mathematical induction. 
$\sum_{i=1}^n i^3 = \biggl(\frac{n(n+1)}{2}\biggl)^2$
My next steps in solving the task:
a)I check the validity of this assertion for n = 1
$\sum_{i=1}^1 1^3 = \biggl(\frac{1(1+1)}{2}\biggl)^2$ = $1$
b)Next, I try to prove the following assertion for n + 1
$\sum_{i=1}^n i^3 = \biggl(\frac{n(n+1)}{2}\biggl)^2$ = $\sum_{i=1}^n i^3$ + $(i + 1)^3$ = $\biggl(\frac{n(n+1)}{2}\biggl)^2$ + $(n+1)^3$ = $\biggl(\frac{n(n+1)}{2}\biggl)^2$ + $(n+1)^3$ = $\biggl(\frac{n(n+1)}{2}\biggl)^2$ + $(n+1)^3$
c)I right solving ? Or have problems ?

You need to show that 
$$
\sum_{i=1}^{n+1} i^3 = \left( \frac{(n+1)(n+2)}{2}\right)^2,
$$
using as an hypothesis that 
$$
\sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2}\right)^2.
$$
So,
\begin{align*}
\sum_{i=1}^{n+1} i^3=& \sum_{i=1}^n i^3 + (n+1)^3=  \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3\\
=& (n+1)^2\left( \frac{n^2}{4}+n+1\right)=\frac{(n+1)^2}{4}(n^2+4n+4)\\
=&\frac{(n+1)^2(n+2)^2}{4} = \left( \frac{(n+1)(n+2)}{2}\right)^2
\end{align*}
