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I've started with the base case first by letting $n=3$, which yields $$LHS = 1^{3}+2^{3}+3^{3} = 36 = \frac{3^{2}\left ( 3+1 \right )^{2}}{4} = RHS.$$

Assume $n=k$ holds: $1^{3}+2^{3}+...+k^{3}=\frac{k^{2}\left ( k+1 \right )^{2}}{4}$ by Induction Hypothesis.

We want to show that $1^{3}+2^{3}+...+\left ( k+1 \right )^{3}=\frac{\left ( k+1 \right )^{2}\left ( k+2 \right )^{2}}{4}.$

I have started from $$1^{3}+2^{3}+...+\left ( k+1 \right )^{3}$$ $$\Rightarrow 1^{3}+2^{3}+...+ k^{3} + 3k^{2} + 3k + 1$$ $$\Rightarrow 1^{3}+2^{3}+...+ + \frac{k^{2}\left ( k+1 \right )^{2}}{4}+3k^{2} + 3k + 1 \text{ (By Induction Hypothesis)}.$$

I am pretty much stuck here so any help would be appreciated.

OGC
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Remark: you can test with $n=3$ but the equality also holds for $n=1$ and $n=2$ so I would consider $n=1$ to be the base case; note that verifying it for $n=1$ is simpler too!

So you assume: $$1^3 + 2^3 + \ldots + k^3 = \frac{k^2(k+1)^2}{4} \tag{1}$$ and you want to show that: $$\color{blue}{1^3 + 2^3 + \ldots + k^3} + (k+1)^3= \frac{(k+1)^2(k+2)^2}{4} \tag{2}$$ Starting from the LHS and using $(1)$ on the blue part, you have: $$\color{blue}{\frac{k^2(k+1)^2}{4}}+ (k+1)^3$$ Now simplify and factor again to show this is equal to the RHS of $(2)$; can you finish?

Hoover over for a start:

$\frac{k^2(k+1)^2}{4}+ (k+1)^3 = \frac{(k+1)^2}{4}\left(k^2+4(k+1) \right) = \ldots$

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