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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?

Arturo Magidin
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Karl
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3 Answers3

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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.

Ross Millikan
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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.

Community
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Bill Dubuque
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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)

alQpr
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  • The reason I thought it was less slick is the x + 1/x substitution leads to y(y^2-3)=52, which (thinking about the factors of 52) makes y=4 pretty easy to spot. No messy square roots involved until you try to get x from y. But it's a matter of taste. – Ross Millikan Jan 05 '11 at 01:06