x*(x-1)(x^2+x+1)=6(x^2+x+1)
Notice that x^2+x+1 is on both sides so divide by it
x*(x-1)=6
x^2-x-6=0
(x-3)(x+2)=0
x= 3
x=-2
For real numbers!
x ( x^3 - 1) - 6(x^2 + x + 1) = 0
Factor (x^3 - 1) as a difference of cubes
x ( x - 1) (x^2 + x + 1) - 6( x^2 + x + 1) = 0
(x^2 + x + 1) [ x(x - 1) - 6 ] = 0
(x^2 + x + 1) [ x^2 - x - 6 ] = 0
(x^2 + x + 1) ( x - 3) ( x + 2) = 0
The first factor doesn't produce any real number roots when set to 0 and solved
The real roots are found by setting the other two factors to 0 and solving for x ⇒ x =3 and x = -2