<s>[INST] 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:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2/3$
$a^2+b^2+c^2=\frac{21}{25}$
$a^3+b^3+c^3=-\frac{194}{125}$
 [/INST] 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
$$
a^3+b^3+c^3 = -\frac{194}{125}
$$
</s>