Question 1119660
<br>
You didn't write the first equation correctly; you left out the linear term.<br>
{{{x^5+x^4+x^3+x^2+x+1 = 0}}}
{{{(x+1)(x^4+x^2+1) = 0}}}
{{{(x+1)((x^4+2x^2+1)-x^2) = 0}}}
{{{(x+1)(((x^2+1)+x)((x^2+1)-x)) = 0}}}
{{{(x+1)(x^2+x+1)(x^2-x+1) = 0}}}<br>
The linear factor gives the only real solution, x = -1.  Each of the quadratic factors gives a pair of complex solutions.<br>
To factor the second expression, we can use a method similar to that used above to factor x^4+x^2+1.<br>
{{{2x^2y^2+2y^2z^2+2x^2z^2-x^4-y^4-z^4}}}
= {{{(-1)(x^4+y^4+z^4-2x^2y^2-2x^2z^2-2y^2z^2)}}}
= {{{(-1)((x^4+y^4+z^4+2x^2y^2-2x^2z^2-2y^2z^2)-4x^2y^2)}}}
= {{{(-1)((x^2+y^2-z^2)^2-(2xy)^2)}}}
= {{{(-1)(x^2+y^2-z^2+2xy)(x^2+y^2-z^2-2xy)}}}