How Can You Rewrite x²/(x⁴+x²+1) Using u?

[anemone]

Here is this week's POTW:

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Given that $u=\dfrac{x}{x^2+x+1}$. Express in terms of $u$ the value of the expression $\dfrac{x^2}{x^4+x^2+1}$.

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[anemone]

Congratulations to the following members for their correct solution::)

1. kaliprasad
2. Theia
3. lfdahl
4. greg1313
5. Opalg

Solution from Theia:

Spoiler

We have

\(\displaystyle u = \frac{x}{x^2 + x + 1}\).

Let's rise both sides to square and one obtains

\(\displaystyle u^2 = \frac{x^2}{x^4 + x^2 + 1 + 2x^3 + 2x^2 + 2x}\).

If one then multiplies the denominator to left hand side and groups the terms, one obtains

\(\displaystyle u^2(x^4 + x^2 + 1) + 2xu^2(x^2 + x + 1) = x^2\).

Now one can substitute the \(\displaystyle (x^2 + x + 1)\) to be equal to \(\displaystyle \frac{x}{u}\) by using the original equation. Thus the expression simplifies to

\(\displaystyle u^2(x^4 + x^2 + 1) + 2x^2u = x^2 \quad \Leftrightarrow \quad u^2(x^4 + x^2 + 1) = x^2(1 - 2u) \),

and further be dividing both sides by \(\displaystyle (x^4 + x^2 + 1)(1 - 2u)\)

\(\displaystyle \frac{x^2}{x^4 + x^2 + 1} = \frac{u^2}{1 - 2u}\).

Alternate solution from Opalg:

Spoiler

Let $u = \dfrac x{x^2+x+1}, \quad v = \dfrac x{x^2-x+1}$. Then $$\frac1u = x + 1 + \frac1x,\qquad \frac1v = x - 1 + \frac1x.$$ Therefore $\dfrac1v = \dfrac1u - 2 = \dfrac{1-2u}u,$ so that $v = \dfrac u{1-2u}.$

It follows that $\dfrac{x^2}{x^4 + x^2 + 1} = \dfrac{x^2}{(x^2 + x + 1)(x^2 - x + 1)} = uv = \dfrac{u^2}{1-2u}.$