Question 763381
This is very similar to question 763384, which has been solved.

We will use the identity 

{{{(a + b)^3 = a^3 + b^3 + 3*a*b(a + b)}}}  ------- (1)

{{{x + 1/x = 4}}}

{{{(x + 1/x)^3 = 4^3 = 64}}}

But by the identity (1),

{{{(x +  1/x)^3 = x^3 + 1/x^3 + 3*x*(1/x)*(x + 1/x)}}} ------- (2)

Substituting for (x + 1/x) as 4

{{{(x + 1/x)^3 = x^3 + 1/x^3 + 3*1*4}}} ------- (3)   ---> x*1/x = 1

{{{64 = x^3 + 1/x^3 + 12}}}    ------ (4)

So, 

{{{x^3 + 1/x^3 = 64 - 12 = 52}}}

:)