I want to find the root of $x^3 - 6 \sqrt{3}=0$ using the bisection method (implemented thru C++) and I found it to be approximately $2.18244...$ but the number of iterations is 36. Is it possible to rewrite the orginal equation so that there are fewer number of iterations?
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3It's possible to use a better method than bisection, e.g., Newton's Method. – Gerry Myerson Mar 13 '23 at 01:43
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Also, perhaps ironically, to in-effect treat $\sqrt{3}$ with infinite precision, consider the equation $(x^3-6\sqrt{3})(x^3+\sqrt{3})=x^6-108=0$? :) – paul garrett Mar 13 '23 at 01:46
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1@paulgarrett You meant $(x^3 -6\sqrt 3 )(x^3 +\color{blue}{6} \sqrt 3) = x^6-108$. – Rócherz Mar 13 '23 at 01:53
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You have $x^3-a^3=(x-a)(x^2+ax+a^2)$ where $a^3=6\sqrt3=2\times 3\sqrt 3=2(\sqrt 3)^3$. – Rócherz Mar 13 '23 at 01:59
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@Rócherz, oops, yes, typo, thanks. :) – paul garrett Mar 13 '23 at 13:31