Question 1000619
set k = 13 and try x = 1

Solve for x over the real numbers:
x^3 -6x^2 +13x -8 = 0

1 -6 +13 -8 = 0

now use synthetic division to find the other factor, (x-1) being one factor

1 | 1 -6 13 -8 |
  |    1 -5  8 |
  | 1 -5  8  0 |

Our two factors are:
(x-1) (x^2 -5x +8)

Split into two equations:
x-1 = 0 or x^2 -5x +8 = 0

Add 1 to both sides:
x = 1 or 

x^2 -5x +8 = 0

Subtract 8 from both sides:
x = 1 or x^2 -5x = -8

Add 25/4 to both sides:
x = 1 or x^2 -5x +25/4 = -7/4

Write the left hand side as a square:
(x -5/2)^2 = -7/4

(x-5/2)^2 = -7/4 has no solution since for all x on the real line, (x-5/2)^2 >=0 and -7/4<0:

Answer:  x = 1  and k = 13