# Roots

Solve for $$r:$$ $$\frac{1}{r}$$+$$\frac{1}{r^{2}}$$+$$\frac{1}{r^{3}}$$+$$\frac{1}{r^{4}}$$$$=340$$

I was only able to solve $r=\frac{1}{4}$ but WolframAlpha has $r ≈ -0.221$, which I don't know how to get.

Note by Ryan Merino
1 month, 1 week ago

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- 1 month, 1 week ago

Neat! I haven't worked much with cubics, obviously. I didn't understand the end of the second example though:

$y = s-t = \sqrt[3]{10 + 6\sqrt{3}} - \sqrt[3]{-10 + 6\sqrt{3}} = 2$

Is there something I'm missing?

- 1 month, 1 week ago

$\dfrac8{t^3} - t^3 = 20$

$(\tfrac2t - t)^3 + 3(\tfrac2t - t)\cdot \tfrac2t \cdot t = 20$

$(\tfrac2t - t)^3 + 6(\tfrac2t - t) = 20$

Let $z = \tfrac2t - t$, then $z^3 + 6z = 20$

Rational root theorem tells us that $z=2$ is a root and the other two roots are not real.

so $z = \tfrac2t - t = 2$ only.

Or $s - t = 2$ only.

- 1 month, 1 week ago

Interesting...the wiki didn't include any of those steps. Would you mind detailing how you went from the first equation to the second? Thanks!

- 1 month, 1 week ago

Use $a^3 - b^3 = (a - b)^3 + 3ab(a - b)$

- 1 month, 1 week ago

Cool. Thanks!

- 1 month, 1 week ago

Thanks. Didn't know this method 'til now.

- 1 month ago

Hmm... seems complicated. :)

- 1 month, 1 week ago

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