If k is a constant what is the value of k such that the polynomial k^2 x^3-6kx+9 is divisible by x-1

We have that
k² x³-6kx+9

we know that

The Remainder Theorem says: If the polynomial f(x) is divisible by a bynomial (x-a) then the number "a" is a root of the polynomial f(x)
so
x=1  is a root of the polynomial
therefore
for x=1
k² x³-6kx+9=0> k² (1)³-6k(1)+9=0> k²-6k+9=0

using a graph tool> to resolve the second order equation

the solution is 
k=3
k=3
k²-6k+9=0> (k-3)²=0

the answer is 
k=3

(1/x + 1)^2 + 6 (1/x + 1) +5