This is by no means "precalculus", and no calculator will help you to see that. Also, your $y = 0$ is nonsensical, except if you want to state the problem as "$a, b, c$ are roots to $y(x) = 0$, where $y(t) = 5t^3 ...$"
A simple application of Viete's relations and Newton's formulae solves most of such problems. For instance,
$\sum \frac{1}{a} = \frac{ab + bc + ca}{abc} = \frac{-2/5}{-3/5} = \frac{2}{3}$
$a^2 + b^2 + c^2 = (a + b + c)^2 - 2\sum ab = \frac{1}{25} - \frac{2(-2)}{5} = \frac{21}{25}.$
For the third one, try summing like $y(a) + y(b) + y(c) = 0$ and conclude something about the LHS.
Edit: Of course, you're talking about the Fundamental theorem of Symmetric Polynomials, stating that expressions such as yours (i.e., symmetric) can always be written in terms of the elementary symmetric polynomials appearing in Viete's relations.