The value which is not possible for $\sum_{k=1}^n k^3$ is 
a)25
b)36
c)225
d)441

The value becomes $$\left[\frac{(n)(n+1)}{2}\right]^2$$
Since all options are perfect squares, I don’t really know what to do with This.
The discrimaint ie. $b^2-4ac>0$ for all options if we solve the quadratic equation for each option.
The right answer is $25$. How do we get that?

$\sum_{k=1}^{n}k^3$ is always the square of a triangular number, and $m$ is a triangular number iff $8m+1$ is a square. It turns out that options b,c,d are fine (they correspond to $n=3,5,6$), but a is not since $8\cdot 5+1$ is not a square.
