To try on: n=1 then \(A=1^3=1;\dfrac{1^2\left(1+1\right)^2}{4}=\dfrac{4}{4}=1\)

Suppose A be true for n = k; that is:

 \(1^3+2^3+......+k^3=\dfrac{k^2\left(k+1\right)^2}{4}\)

We shall show that A is true too

Add \(\left(k+1\right)^3\),we have:

 \(1^3+2^3+........+k^3+\left(k+1\right)^3\)=\(\dfrac{k^2\left(k+1\right)^2}{4}+\left(k+1\right)^3=\dfrac{k^2\left(k+1\right)^2+4\left(k+1\right)^3}{4}\)

\(=\dfrac{\left(k+1\right)^2.\left(k^2+4k+4\right)}{4}=\dfrac{\left(k+1\right)^2.\left(k+2\right)^2}{4}\)

So the supposition right