Question 760067
{{{(3x)^2=3^2*x^2=9x^2}}} so {{{x^3+(3x)^2-7x+1=0}}} means {{{x^3+9x^2-7x+1=0}}}
You may have meant {{{x^3+3x^2-7x+1=0}}}, where only the {{{x}}} is squared.
 
THE PROBLEM AS POSTED:
Let the solutions of {{{x^3+(3x)^2-7x+1=0}}} be {{{a}}}, {{{b}}}, and {{{c}}}.
The full factorization of the polynomial {{{x^3+(3x)^2-7x+1}}} must be
{{{x^3+(3x)^2-7x+1=(x-a)(x-b)(x-c)}}}
If we multiply that factorization we get
{{{x^3+(3x)^2-7x+1=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc}}}
From that, we get
{{{-(a+b+c)=9}}} --> {{{a+b+c=-9}}} and
{{{ab+ac+bc=-7}}}
If we multiply, we find that
{{{(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)}}}
Replacing the values found for the polynomial,
{{{(-9)^2=a^2+b^2+c^2+2(-7)}}} --> {{{81=a^2+b^2+c^2-14}}} --> {{{a^2+b^2+c^2=81+14}}} --> {{{a^2+b^2+c^2=95}}}  
 
IF THE PROBLEM EQUATION IS {{{x^3+3x^2-7x+1=0}}}:
Let the solutions of {{{x^3+3x^2-7x+1=0}}} be {{{a}}}, {{{b}}}, and {{{c}}}.
The full factorization of the polynomial {{{x^3+(3x)^2-7x+1}}} must be
{{{x^3+3x^2-7x+1=(x-a)(x-b)(x-c)}}}
If we multiply that factorization we get
{{{x^3+3x^2-7x+1=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc}}}
From that, we get
{{{-(a+b+c)=3}}} --> {{{a+b+c=-3}}} and
{{{ab+ac+bc=-7}}}
If we multiply, we find that
{{{(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)}}}
Replacing the values found for the polynomial,
{{{(-3)^2=a^2+b^2+c^2+2(-7)}}} --> {{{9=a^2+b^2+c^2-14}}} --> {{{a^2+b^2+c^2=9+14}}} --> {{{a^2+b^2+c^2=23}}}