$1^3$ + $2^3$ + $2^3$ + ... + $n^3$ = ($1 + 2 + 3 + ... + n)^2$
I start with $P(1)$ and get $1 = 1$.
Then I do it with $P(n+1)$ and I get stuck.
$1^3$ + $2^3$ + $2^3$ + ... + $n^3$ + $(n+1)^3$ = ($1 + 2 + 3 + ... + n +(n+1))^2$
then I've tried substituting values and both ways and I cannot find anywhere to go with the problem.
$(1 + 2 + 3 + ... + n)^2 + (n+1)^3 = (1 + 2 + 3 + ... + n +(n+1))^2$ $OR$
$1^3 + 2^3 + 2^3 + ... + n^3 + (n+1)^3 = (1^3 + 2^3 + 2^3 + ... + n^3 +(n+1))^2$
I know that the sum of numbers in a row is $\cfrac{n(n+1)}{2}$. I'm not sure if that's of any use. I'm pretty sure there's a mathematical proof that I can't remember that'll clean up this problem so any insight would be greatly appreciated.
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