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I was solving a cubic equation with two different methods and one of the root I got with first method is $\frac{1}{\omega-3\omega^2}$ and with second method I got $\frac{4\omega^2+3}{13}$.

$\omega$ is the cube root of unity

If I set one equals another then I got an equality $13=13$ which indicates that they are equal.

$$\frac{1}{\omega-3\omega^2}=\frac{\omega^2}{1-3\omega}$$

After this I think we have to do rationalization kind of thing to the make denominator real but I could not figure out How.

Thank you for your help!

user6262
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    You can multiply by the conjugate of the denominator to make it real. – dxiv Feb 05 '22 at 08:44
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    We get $$\frac{1}{\omega-3\omega^2}= \frac{1\cdot \omega^2}{(\omega-3\omega^2)\cdot\omega^2}=\frac{\omega^2}{\omega^3-3\omega^4}=\frac{\omega^2}{1-3\omega}$$ – Jochen Feb 05 '22 at 08:45
  • @Jochen Actually that's not what I wanted to show(I have showed that)I wanted to show problem in title of question – user6262 Feb 05 '22 at 08:59

3 Answers3

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The conjugate of $\omega$ is $\omega^2$, so multiply top and bottom by $\omega^2-3\omega$. This gives

$$\frac{1}{\omega-3\omega^2}$$

$$=\frac{1}{\omega-3\omega^2}\cdot\frac{\omega^2-3\omega}{\omega^2-3\omega}$$

$$=\frac{\omega^2-3\omega}{1-3\omega^2-3\omega^4+9}$$

Use $\omega^3=1$ and $1+\omega+\omega^2=0$ to give

$$=\frac{\omega^2-3\omega}{1-3(\omega^2+\omega)+9}$$

$$=\frac{\omega^2-3\omega}{13}$$

$-3\omega=3+3\omega^2$ then gives your result.

JMP
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As a hint : if $\omega$ is cube root of unity, so $\omega^3=1$ put in like below and simplify $$\frac{1}{\omega-3\omega^2}=\frac{\omega^3}{\omega-3\omega^2}$$

Khosrotash
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You can multiply by the conjugate of the denominator to make it real, but another approach is:
$(\omega-3\omega^2)(4\omega^2+3)=4\omega^3+3\omega-12\omega-9\omega^2=4-9\omega-9\omega^2=13$
Dividing by $13(\omega-3\omega^2)$ gives the eqn in title.

cineel
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