### Instruction: Given that $x$ is real and $x^3+\frac{1}{x^3}=52$, find $x+\frac{1}{x}$.### Response: To solve this problem, I notice that the expression $x^3+\frac{1}{x^3}$ is related to the expression $x+\frac{1}{x}$ by a cubic identity.
Specifically, I recall that $(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3x+\frac{3}{x}$.
If I isolate $x^3+\frac{1}{x^3}$ on one side, I get $x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})$.
This means that I can substitute $52$ for $x^3+\frac{1}{x^3}$ in the given equation and get $52=(x+\frac{1}{x})^3-3(x+\frac{1}{x})$.
To simplify this equation, I let $y=x+\frac{1}{x}$ and rewrite it as $y^3-3y-52=0$.
This is a cubic equation in $y$, which I can try to factor by looking for rational roots.
By the rational root theorem, the possible rational roots of this equation are $\pm 1, \pm 2, \pm 4, \pm 13, \pm 26, \pm 52$.
I can test each of these values by plugging them into the equation and seeing if they make it true.
I find that $y=4$ is a root, since $4^3-3(4)-52=0$.
This means that I can factor out $(y-4)$ from the equation and get $(y-4)(y^2+4y+13)=0$.
The other factor, $y^2+4y+13$, has no real roots, since its discriminant is negative.
Therefore, the only real solution for $y$ is $y=4$.
But $y$ is the same as $x+\frac{1}{x}$, so I have found the answer.