1. You mention the method to be used for the solution, Vieta.
If the roots of the cubic are a, b and c, then
\(\displaystyle (x-a)(x-b)(x-c) = 0, \\ \text{expanding,} \\ x^{3}-x^{2}(a+b+c)+x(ab+bc+ca)-abc=0.\)
So, for the given cubic,
\(\displaystyle a+b+c=3\dots \dots(1) \\ ab+bc+ca=-15/2 \dots\dots(2)\\ abc=3/2\dots \dots (3).\)
It's just follow your nose after that.
\(\displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} =\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}{a^{2}b^{2}c^{2}}.\)
The denominator is easy, just substitute (3).
For the numerator it looks like you have to make use of (2), so,
\(\displaystyle (ab+bc+ca)^{2}=a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2a^{2}bc+2ab^{2}c+2abc^{2}\\= a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc(a+b+c),\)
and now it's just substitution from (1), (2) and (3).
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