Question 162337
equation is {{{3*x^3*y^2-3*x^2*y^2+3*x*y^2}}}
looks like you can factor out 3*x*y^2 to get
{{{3*x*y^2*(x^2 - x + 1)}}}
quadratic formula to find the roots of {{{x^2 - x + 1}}} is
{{{(-b + sqrt(b^2 - 4*a*c))/(2*a)}}} and {{{(-b - sqrt(b^2 - 4*a*c))/(2*a)}}}
looking at {{{x^2 - x + 1}}} and using the quadratic formula, it looks like the solution to that quadratic equation is not real since
the part of the equation under the square root sign {{{b^2 - 4*a*c}}} is negative as shown below.
in the equation {{{x^2 - x + 1}}}
a = 1
b = -1
c = 1
4*a*c = 4
b^2 = 1
{{{sqrt(b^2 - 4*a*c)}}} equals {{{sqrt(1-4)}}} = {{{sqrt(-3)}}} which is negative so you don't have a real solution.
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the answer appears to be
{{{3*x*y^2*(x^2 - x + 1)}}}
{{{x^2 - x + 1}}} appears to be prime because it can't be factored any further.