如何证明这两个方程有一个公共根
\[\tan \frac{\pi }{9} = \frac{{4s\left( {3 - 13{s^2} + 13{s^4} - 3{s^6}} \right)}}{{1 + 28{s^2} - 74{s^4} + 28{s^6} + {s^8}}}\]\
数值显示这两方程有一个公共根: 0.4663076581549985928300061947995594513110630082513759178109324482392183833889065365531268232572728731 辗转相除 哦明白了\[{\rm{Eliminate}}\left[ {\left\{ {{s^8}\left( { - \tan \left( {\frac{\pi }{9}} \right)} \right) - 12{s^7} - 28{s^6}\tan \left( {\frac{\pi }{9}} \right) + 52{s^5} + 74{s^4}\tan \left( {\frac{\pi }{9}} \right) - 52{s^3} - 28{s^2}\tan \left( {\frac{\pi }{9}} \right) + 12s - \tan \left( {\frac{\pi }{9}} \right) = 0,{s^6} - 12{s^5} + 3{s^4} + 40{s^3} + 3{s^2} - 12s + 1 = 0} \right\},s} \right]\]
\^2} - 33\tan {\left[ {\frac{\pi }{9}} \right]^4} + \tan {\left[ {\frac{\pi }{9}} \right]^6} = 3\] s->\(\frac{1}{4}+\frac{1}{4} \tan ^5\left(\frac{\pi }{9}\right)-\frac{1}{4} \tan ^4\left(\frac{\pi }{9}\right)-8 \tan ^3\left(\frac{\pi }{9}\right)+8 \tan ^2\left(\frac{\pi }{9}\right)-\frac{5}{4} \tan \left(\frac{\pi }{9}\right)\)
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