Step 1: Recall the sum of squares formula. The sum $1^2+2^2+\cdots+n^2$ equals $\dfrac{n(n+1)(2n+1)}{6}$. We will compare this with the options. Step 2: Open the brackets. First $(n+1)(2n+1)=2n^2+3n+1$. So the sum is \[ S=\frac{n(2n^2+3n+1)}{6}=\frac{2n^3+3n^2+n}{6}. \] Step 3: Pull out the cube term. Split the fraction: $S=\dfrac{2n^3}{6}+\dfrac{3n^2+n}{6}=\dfrac{n^3}{3}+\dfrac{3n^2+n}{6}$. Step 4: Check the leftover part. The leftover $\dfrac{3n^2+n}{6}$ is always more than zero for every natural number $n$, because both $3n^2$ and $n$ are positive. Step 5: Form the inequality. Since we add a positive amount to $\dfrac{n^3}{3}$, the full sum is always larger than $\dfrac{n^3}{3}$. So $S>\dfrac{n^3}{3}$. Step 6: Rule out the others. Try $n=1$: the sum is $1$. Then $\dfrac{n^3}{2}=0.5$ and $n^3=1$ and $3n^3=3$. Only $\dfrac{n^3}{3}\approx0.33$ stays below the sum for all $n$. So option 3 is correct. \[ \boxed{1^2+2^2+\cdots+n^2>\frac{n^3}{3}} \]