Question

A) x^2-2x+1=0
b) x^3-3x^2x+3x-1=0
Find x.

Answer 1

Part (a)

`(x-1)^2=0 \implies x=\boxed{1}`

Part (b)

`x^3-3x^2 x+3x-1=0 \\ \\ -2x^3+3x-1=0`

By inspection, x=1 is a root, so we can consider dividing by x-1.

`(-2x^3+3x-1)/(x-1)=-2x^2-2x+1 \\ \\ \implies (x-1)(-2x^2-2x+1)=0 \\ \\ x-1=0 \implies x=\boxed{1} \\ \\ -2x^2-2x+1=0 \implies x=(2 \pm √(4+8))/(-4) \\ \\ =\boxed{(-1 \pm √(3))/(2)}`