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I am trying to prove that for any integer where $n \ge 1$, this is true:

$$ (1 + 2 + 3 + \cdots + (n-1) + n)^2 = 1^3 + 2^3 + 3^3 + \cdots + (n-1)^3 + n^3$$

I've done the base case and I am having problems in the step where I assume that the above is true and try to prove for $k = n + 1$.

I managed to get,

$$(1 + 2 + 3 + \cdots + (k-1) + k + (k+1))^2 = (1 + 2 + 3 + \cdots + (k-1) + k)^2 + (k + 1)^3$$

but I'm not quite sure what to do next as I haven't dealt with cases where both sides could sum up to an unknown integer.

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Hint: It might help to prove by induction that $1+2+\cdots+n = \dfrac{n(n+1)}{2}$ first, then use this form to prove $1^3 + 2^3 + \cdots + n^3 = \dfrac{n^2(n+1)^2}{4}$

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