Really stuck on strategy for this fundamental theorem of algebra type problem. I am at a loss at how to approach this problem. It doesn't make sense to me to solve without finding the roots. I couldn't find the roots without a calculator anyway, but you're not meant to.
Show that if the roots of the equation:
$5x^3-x^2-2x+3=0$
are a,b and c then:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2/3$
$a^2+b^2+c^2=\frac{21}{25}$
$a^3+b^3+c^3=-\frac{194}{125}$
 A: 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.
A: The hint referring to the Fundamental Theorem of Algebra was to get you to realize that a cubic polynomial will have three complex roots. In other words, we can write your polynomial as 
$$
y=x^3-\frac{1}{5}x^2-\frac{2}{5}x+\frac{3}{5}
$$ 
and as
$$
y=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc
$$
so equating terms with like exponents we have

$a+b+c=\frac{1}{5}$
$ab+ac+bc=-\frac{2}{5}$
$abc=-\frac{3}{5}$

Now we can solve your problems.

(1) Using the values above we have 
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{ab+ac+bc}{abc} = \frac{-2/5}{-3/5}=\frac{2}{3}
$$

(2) From
$$
(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)
$$
we have
$$
\left(\frac{1}{5}\right)^2 = (a^2+b^2+c^2)+2\left(-\frac{2}{5}\right)
$$
so $a^2+b^2+c^2=21/125.$

(3) It'll be easiest to use a preliminary result for this part. From
$$
(a+b+c)(ab+ac+bc) = (a^2b+ab^2+a^2c+ac^2+b^2c+bc^2) + 3abc
$$
it's not hard to show that
$$
a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=\frac{43}{25}
$$
So from
$$
(a+b+c)^3 = (a^3+b^3+c^3)+(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+2(a+b+c)(ab+ac+bc)
$$
we can conclude
$$
a^3+b^3+c^3 = -\frac{194}{125}
$$
A: If you haven't yet been exposed to the general theory, there's nothing stopping you working from first principles:
Given
$$
5x^3−x^2−2x+3=0
$$
we have
$$
x^3−(1/5)x^2−(2/5)x+3/5=0
$$
Now if $a$, $b$, $c$ are the roots, then
$$
(x-a)(x-b)(x-c) = 0
$$
which is to say
$$
x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0
$$
so we know
$$
\begin{align*}
a+b+c &= 1/5\\ 
ab+bc+ca &= -2/5\\
abc &= -3/5
\end{align*}
$$
which are enough to establish the things you're asked to establish, with some manipulation. For example,
$$
\begin{align*}
1/a + 1/b + 1/c &= bc/abc + ca / abc + ab / abc\\
 &= (ab+bc+bc) / abc
\end{align*}
$$
and
$$
a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)
$$
The sum of cubes is left as an exercise.
