# Rationalize the denominator with 4 terms $\frac{1}{1+\sqrt{12}-\sqrt[3]{3}-\sqrt[3]{9}}$

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Your expression "simplifies" to $${-3\cdot3^{1/6} + 3^{1/3} + 7 \sqrt{3} + 5\cdot3^{2/3} + 2\cdot3^{5/6} - 2\over 22}$$ Outline of solution: First, transform the given expression to this convenient form: $${1\over(1-3^{1/3}+3^{1/6})(1+3^{1/3}-3^{1/6})},$$ then multiply both the numerator and denominator by $(3^{1/3}+3^{1/6}+1)(3^{1/3}+3^{1/6}-1)$ so that the denominator looks more symmetric, like Heron's formula: $$(3^{1/3}+3^{1/6}+1)(3^{1/3}+3^{1/6}-1)(1-3^{1/3}+3^{1/6})(1+3^{1/3}-3^{1/6})$$ Simplify the denominator (let $x=3^{1/6}$): $$(x^2+x+1)(x^2+x-1)(1-x^2+x)(1+x^2-x) = 2x^6-1-(x^6-2)x^2+x^4$$ $$= 5-3^{1/3}+3^{2/3} \quad\mbox{(here we used: }\ 2x^6-1=5, \ \ (x^6-2)x^2=3^{1/3}).$$ So the denominator becomes $5-3^{1/3}+3^{2/3}$. Finally, use the fact that $$(5 - 3^{1/3} + 3^{2/3})\times(7 + 2\cdot3^{1/3} - 3^{2/3}) = 44.$$ Note: This is a special case ($n=3$, $x=3^{1/6}$) of the following: for $x=n^{1/6}$, $n\in{\mathbb N}$, the product $$((2n-1)-(n-2)x^2+x^4) \ \times \ ((5n-1)+(2n-2)x^2+(n-5)x^4)$$ $$= -(n - 1) (n^2 - 18 n + 1)$$ is an integer.

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Updated on December 01, 2022

$$\frac{1}{1+\sqrt{12}-\sqrt[3]{3}-\sqrt[3]{9}}$$
And your original reply to my warning was ** I want to see how many and how quickly downvotes come**, Atleast now you know how quickly they come. Anyways Happy New Year $\displaystyle \huge \ddot \smile$.