when u said
x(x-3)(x-1)=kx
either x=0 or x=something else
(x-3)(x-1)=k if x isn't 0
when k=3,
(x-3)(x-1)=3
x=0 or x=4,
but you mentioned that x isn't zero so the solution is x=4 only
or from original eqn,
x(x-3)(x-1)=kx
x^3 - 4x^2 + 3x = kx
when k=3,
x^3 - 4x^2 + 3x = 3x
x^3 -...