Question 808593
let the number be x

x + (6/x) = x^3

(x^2+6)/x=x^3

x^2+6=x^4

x^4-x^2-6=0

x^4-3x^2+2x^2-6=0

x^2(x-3)+2x(x-3)=0

(x^2-3)(x^2+2x)=0
x(x^2-3)(x+2)=0

x=0, x= +/- sqrt(3) x=-2

You can find the squares