Need help with a power equation If $x^3 + \frac{1}{x^3} = 52$, then what is $x^2 + \frac{1}{x^2}$?
I'm not sure which formulae or methods should I used to solve this problem, so could somebody show me a way? what should I be looking for?
 A: Hint: Define $y=x+\frac{1}{x}$.  There are frequently problems where this transformation helps.  The equation for $y^3$ has only one real root, which you can probably find by inspection.  Then the quadratic formula will give you the other two.  The problem isn't as clean as I thought originally.
Added:  this works with symmetric polynomials as well.  Imagine you were given $x^6-52x^3+1=0$.  You might be scared by the sixth degree, but if you notice the coefficients are symmetric around the center, you can divide by $x^3$, define $y=x+\frac{1}{x}$, and cut the degree in half.  Another approach, which doesn't work as slickly on the current problem, is to define $u=x^3$ and rewrite it as $u+\frac{1}{u}=52$.  This is a quadratic which will yield to the usual techniques.  I have seen cases where people give you what looks like a high degree polynomial, but this will tame them.
A: HINT $\ \ $ Exploit the symmetry. $\ $ Put $\rm\quad x_k = x^k + x^{-k}\:.\ $ Then we have  
$\rm\quad\quad\ x_1^2 = (x+1/x)^2 = x_2 + 2\quad\quad\ $ so $\rm\ \ x_2 = x_1^2 - 2\quad\quad\quad\quad\quad(2)$
$\rm\quad\quad\ x_1^3 = (x+1/x)^3 = x_3 + 3\ x_1\ $ so $\rm\quad x_3 = x_1^3 - 3\ x_1\quad\quad\quad (3)$
Thus, put $\rm\:x_3 = 52\:$ in $(3)$, solve it for $\rm\:x_1\:,\:$ then plug $\rm\:x_1\:$ into $(2)$ to get $\rm\:x_2\:$.
Analogous methods work for arbitrary symmetric polynomials, e.g. see here.
A: After clearing the fraction (multiplying through by x^3), the given equation is a quadratic in x^3. Solve to get x^3=26+/-15sqrt(3), then just take the cube root(s) to get x, and plug in.
(I am not sure why Ross Millikan considers this less "slick" than any other method)
