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Homework Statement
Find exact values for z without any trigonometrical components or imaginary numbers
[tex]x = \frac{79}{60} + \frac{1}{30} \sqrt{6121} cos\left(\frac{1}{3}arccos(z)\right) [/tex]
where
[tex]z = \frac{473419}{6121\sqrt{6121}} [/tex]
Homework Equations
[tex]cos\left(\frac{1}{3}arccos(z)\right) = \frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}} [/tex]
The Attempt at a Solution
I pretty fast get imaginary numbers from the equation above. But they shall, as I know, cancel one way or the other. So this is what I get:
[tex]x = \frac{79}{60} + \frac{1}{30} \sqrt{6121} cos\left(\frac{1}{3}arccos(z)\right) = \frac{79}{60} + \frac{1}{30} \sqrt{6121} \frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}}[/tex]
[tex]x =
\frac{79}{60} + \frac{1}{30} \sqrt{6121} \left(\frac{\left({ \frac{473419}{6121\sqrt{6121}} + \sqrt{ -\frac{5207760000}{229333309561}}}^{1/3}}{2} +
\frac {1}{2\left({ \frac{473419}{6121\sqrt{6121}} + \sqrt{ -\frac{5207760000}{229333309561}}}\right)^{1/3}}\right)[/tex][tex]x =
\frac{79}{60} + \frac{1}{60} \sqrt{6121} \left(\left({ \frac{473419}{6121\sqrt{6121}} + i \sqrt{\frac{5207760000}{229333309561}}}\right)^{1/3} +
\frac 1 {\left({ \frac{473419}{6121\sqrt{6121}} + i \sqrt{\frac{5207760000}{229333309561}}}\right)^{1/3}}\right)[/tex][tex]x =
\frac{79}{60} + \frac{1}{60} \left({473419 + 600 i \sqrt{14466}}\right)^{1/3} +
\frac {6121}{60\left({473419 + 600 i \sqrt{14466}}\right)^{1/3}}[/tex]
Let
[tex]u = 473419 + 600 i \sqrt{14466}[/tex]
Then I have
[tex]x =
\frac{79}{60} + \frac{1}{60} {u}^{1/3} +
\frac {6121}{60{u}^{1/3}}[/tex]
[tex]x =
\frac{79}{60} + \frac{{u}^{2/3} + 6121}{
{60{u}^{1/3}}}[/tex]
[tex]x =
\frac{79}{60} + \frac{{u}^{4/3} + 6121{u}^{2/3}}{
{60u}}[/tex]But from here, I can not see how I can get rid of the i-s...
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