x^3-7x+6 factorize by vanishing method
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x^3-7x+6 = 0
x^3-(6+1)x+6 = 0
x^3 - 6x - x + 6 = 0
x^2(x-6)-1(x-6) = 0
(x-6) (x^2-1) = 0
EITHER, x-6=0 => x = 6
OR, x^2-1=0 => x^2=1 => x = √1
x^3-(6+1)x+6 = 0
x^3 - 6x - x + 6 = 0
x^2(x-6)-1(x-6) = 0
(x-6) (x^2-1) = 0
EITHER, x-6=0 => x = 6
OR, x^2-1=0 => x^2=1 => x = √1
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