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Now $\dfrac{x}{x^2+x+1}=a$ is known, I want to express the $\dfrac{x^2}{x^4+x^2+1}$ in the forms of a, how should I do it?

The answer is $\dfrac{a^2}{1-2a}$, and there is a limitation of $-1\leq a \leq \dfrac{1}{3}$, of course.

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  • $\begingroup$ Another function Eliminate: Eliminate[{y == x^2/(x^4 + x^2 + 1),x/(x^2 + x + 1) == a},x] gives a^2 + 2 a y == y. $\endgroup$
    – lilyric
    Aug 15 at 2:34
  • $\begingroup$ @lilyric Thanks. Then Solve[a^2 + 2 a y == y, {y}, Reals] will give the form I want. $\endgroup$ Aug 15 at 2:38

1 Answer 1

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We can solve y and eliminate x.

Solve[{y == x^2/(x^4 + x^2 + 1), 
   x/(x^2 + x + 1) == a}, {y}, {x}] // FullSimplify

enter image description here

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  • $\begingroup$ It throws a Solve::fulldim error which instructs me to use Reduce, while the result is {{}}. $\endgroup$ Aug 15 at 2:31
  • $\begingroup$ @ThomasPeng Clear[x, y, a]; Solve[{y == x^2/(x^4 + x^2 + 1), x/(x^2 + x + 1) == a}, y, {x}] // Simplify $\endgroup$
    – cvgmt
    Aug 15 at 2:35
  • $\begingroup$ @ThomasPeng Clear[x, y, a]; Solve[{y == x^2/(x^4 + x^2 + 1), x/(x^2 + x + 1) == a}, y, {x}, Method -> Reduce] // Simplify $\endgroup$
    – cvgmt
    Aug 15 at 2:36
  • $\begingroup$ Thanks a lot. I forgot to clear the variables. $\endgroup$ Aug 15 at 2:37

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