Spy's thread.

(the quadratic formula is as follows: negative b plus or minus the square root of b squared minus 4ac all over 2a)

(One can use the quadratic formula to conclude that the cube root of one is negative .5 +- radical three over two times i.
x^3 = 1
X^3 - 1 = 0
(x^2 + x + 1)(x-1) = 0
x^2 + x + 1 = 0
x = (-1+-(1-4)^.5)/2
x= (-1 +- 3^.5 * i)/2
x = -.5 +- (3^.5*i)/2)

(One can use the quadratic formula to conclude that the cube root of one is negative .5 +- radical three over two times i.
x^3 = 1
X^3 - 1 = 0
(x^2 + x + 1)(x-1) = 0
x^2 + x + 1 = 0
x = (-1+-(1-4)^.5)/2
x= (-1 +- 3^.5 * i)/2
x = .5 +- (3^.5*i)/2)
(you divided each side by (x-1), but that is impossible beacuse since x cubed is one, x is one, and 1-1=0. you can't divide by 0

(One can use the quadratic formula to conclude that the cube root of one is negative .5 +- radical three over two times i.
x^3 = 1
X^3 - 1 = 0
(x^2 + x + 1)(x-1) = 0
x^2 + x + 1 = 0
x = (-1+-(1-4)^.5)/2
x= (-1 +- 3^.5 * i)/2
x = -.5 +- (3^.5*i)/2)
(you divided each side by (x-1), but that is impossible beacuse since x cubed is one, x is one, and 1-1=0. you can't divide by 0
(x does not have to be one, however.
I will show that my expression cubed is equal to one. I will only show the positive solution (if you want to verify the negative, do it on your own).
The expression for (a+b)^3 is a^3 + 3a^2*b + 3ab^2 + b^3
a = -.5
b= 3^.5*i/2
x = ((-.5)^3) + (3 * (-.5)^2 * 3^ .5 *i/2) + (3 * -.5 * (3^.5*i/2)^2) + ((3^.5*i/2)^3)
x = -.125 + .6495190528i + 1.125 + -.6495190528i
(The ugly decimal is (-3*3^.5*i)/8)
x = (-.125 + 1.125) + (.6495190528-.6495190528)
x = 1+0
x = 1
)
In here, we all will write anything in white font so that we highlight it. Please put (parentheses) around writing so we know where to go.
(I'm amazed at how grandmasters play. They make moves that seem ridiculous and win with a new move. How do they do it? Thoughts?)