i found proof that 3= 0

[graceful relic]

you assume there's a real x that satisfies x^2+x+1=0 to begin with

this is a geometric series, $$\frac{x^3-1}{x-1}=0$$

[neat sentinel]

Merosity

[graceful relic]

in other words, x^3-1 = (x-1)(x^2+x+1), it's only true that x-1 divides x^3-1, but it doesn't divide x^2+x+1

you can see that the roots of that are imaginary by looking at the quadratic formula, the discriminant is -3

[jovial jolt]

Thanks

Cause I started questioning my life’s choices