If x (x^2+6x+18)(x+6) = 159919 then what would be (x+4)^4

Question

x, y, z and k are four non zero positive integers satisfying 1/x + y/2 = z/3 + 4/k, minimum integral value of k for integral value of x, y and z will be

Answer 1

x(x^2+6x+18)(x+6)=409*23*17
On equating the above equation , we get
x=17.
Therefore ,
(17+4)^4=194481

Answer 2

we have
x (x^2+6x+18)(x+6) = 159919

(x+3-3)(x^2+6x+18)(x+3+3) = 159919
[(x+3)^2 - 9] [(x+3)^2 + 9] = 159919
(x+3)^4 - 81 = 159919
(x+3)^4 = 160000
x+3 = 20
x = 17

therefore,
(x+4)^4= (17+4)^4
= 21^4
= 194481