Answer
$P(n)$ is true for all values of $n$.
Work Step by Step
Suppose that $P(n)=1^3+2^3+3^3+.......+n^3=\dfrac{n^2(n+1)^2}{4}$
1. Our aim is to find that $P(n)$ is true for $n=1$
$1^3=\dfrac{1^2(1+1)^2}{2} \implies 1=1$
So, it is true for $n=1$.
2. Our aim is to find that $P(n)$ is true for $n=k$.This, it will also true for $n=k+1$
$1^3+2^3+3^3+.......+k^3=\dfrac{k^2(k+1)^2}{4}\\ \implies \dfrac{k^2(k+1)^2}{4} +(k+1)^3=\dfrac{(k+1)^2(k+2)^2}{4}$
This yields: $ \dfrac{(k+1)^2(k+2)^2}{4}=\dfrac{(k+1)^2(k+2)^2}{4}$
So, it is true for $n=k+1$.
Therefore, $P(n)$ is true for all values of $n$.
