<s>[INST] Using 3 equations, find $2a-b+c$. From the problem, I extract the following 3 equations.


*

*$a+b+c=10$

*$ab+bc+ca=31$

*$abc=30$


The question is to find $(2a-b+c)$.

Using the equation $(a+b+c)^2 = a^2 + b^2 + c^2 +2 (ab+bc+ca)$, I found $a^2+b^2+c^2=38$.
Using the equation $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2 + b^2+c^2-ab-bc-ca)$, I found $a^3+b^3+c^3=160$.
However, I was stuck here. Please let me know some directions. Thank you.
 [/INST] By Viete $a$, $b$ and $c$ are roots of the following equation:
$$x^3-10x^2+31x-30=0$$ or
$$(x-2)(x-5)(x-3)=0$$
and the rest for you.
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