# 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?

-
$$x^3+\frac1{x^3}=\left(x+\frac1{x}\right)\left(x^2+\frac1{x^2}-1\right)$$ –  non-expert Jan 4 '11 at 14:11

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.

-

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.

-