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:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2/3$
$a^2+b^2+c^2=\frac{21}{25}$
$a^3+b^3+c^3=-\frac{194}{125}$

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.
