Prove $1^3 + 2^3 + \ldots n^3 = \frac{n^2(n + 1)^2}{4}$ with mathematical induction Hello guys I'm trying to prove this equation with mathematical induction method.



So for n = 1 I know that :



1 = 1
Now I know that it works for some k



I wanna prove it for k + 1:



Since :


I can rewrite it as :



So I'm stuck here.. I don't know what should I do.. Should I first square (k + 1)^2 or rewrite (k + 1)^3 ??'
Either way I get weird result and don't know how to proceed with this next step to prove mathematical
induction. I mean I can multiple k^2 with the first parenthesis but I think I should leave it that way and do smth other than that. I'll appreciate any help. Thanks :)



 A: You are almost there! Notice that
\begin{align*}
\frac{k^{2}(k+1)^{2}}{4} + (k+1)^{3} & = \frac{k^{2}(k+1)^{2}+4(k+1)^{3}}{4}\\\\
& = \frac{(k^{2}+4(k+1))(k+1)^{2}}{4}\\\\
& = \frac{(k^{2} + 4k + 4)(k+1)^{2}}{4}\\\\
& = \frac{(k+2)^{2}(k+1)^{2}}{4}\\\\
& = \frac{(k+1)^{2}(k+2)^{2}}{4}\\\\
& = \frac{(k+1)^{2}((k+1)+1)^{2}}{4}
\end{align*}
and we are done.
Hopefully this helps!
A: Note:  It's enough to prove $\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4 = n^3$.
That is to say, if we assume Equation 1: $1^3 + 2^3 + ...... + (n-1)^3=  \frac {(n-1)^2n^2}4$
And we want to prove Equation 2: $1^3 + 2^3 +.... + (n-1)^3 + n^3 = \frac {n^2(n+1)^2}4$
That is a matter of proving by subtraction Equation 2 for Equation 1, that
$n^3= \frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4$
So... Lets do that.
$\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4=$
$\frac {n^2}4[ (n+1)^2 - (n-1)^2] =$
$\frac {n^2}4[\{(n+1)+(n-1)\}\{(n+1)-(n-1)\}]=$
$\frac {n^2}4[\{2n\}\{2\}]=$
$\frac {n^2}4\cdot 4n = n^3$.
And ... that's that.
....
If somehow that doesn't feel right..... you can expand...
$\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4=$
$\frac {n^2 (n^2 + 2n + 1)}4 -\frac {n^2(n^2 -2n+1)}4 = $
... and, yeah, yeah, we get it.
