Rational equation

Question

my question

İf $ x+\frac{1}{x^2}=3$ then find $( x^2 -\frac{1}{x})^2 $ . I tried factoring, taking squares of both sides and some other things that did not work. what should i do?

Answer 1

I did it the hard way: Solve $x^3-3x^2+1=0$ which has three real roots $x_1,x_2,x_3$ in $(-1,3).$ Here are the closed forms computed with Maple:

$$x_1 = -\cos(\frac{\pi}{9})+1-\sqrt{3}\sin(\frac{\pi}{9}) \approx -0.532$$ $$x_2 = -\cos(\frac{\pi}{9})+1+\sqrt{3}\sin(\frac{\pi}{9}) \approx 0.653$$ $$x_3 = 2\cos(\frac{\pi}{9})+1 \approx 2.879$$

Then compute $(x_i^2-\frac{1}{x_i})^2.$ You can imagine that the closed forms are even more complicated.

Answer 2

Contrary to $x+\frac{1}{x^2}=3$, the expresion $x-\frac{1}{x^2}$ cannot be constant because if $(x-\frac{1}{x^2})^2=a^2$ then one would have $$x^3-3x^2+1=0\\x^3\mp ax^2-1=0$$ from which $$x=\pm\sqrt{\frac{2}{3\mp a}}$$ and this is not compatible with the three distinct possible (real) values of $x$.

An easy calculation gives $$\left(x-\frac{1}{x^2}\right)^2=\left(3-\frac{2}{x^2}\right)^2$$