Rationalize the denominator with 4 terms $\frac{1}{1+\sqrt{12}\sqrt[3]{3}\sqrt[3]{9}}$
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: $$ {1\over(13^{1/3}+3^{1/6})(1+3^{1/3}3^{1/6})}, $$ 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: $$(3^{1/3}+3^{1/6}+1)(3^{1/3}+3^{1/6}1)(13^{1/3}+3^{1/6})(1+3^{1/3}3^{1/6}) $$ Simplify the denominator (let $x=3^{1/6}$): $$(x^2+x+1)(x^2+x1)(1x^2+x)(1+x^2x) = 2x^61(x^62)x^2+x^4 $$ $$= 53^{1/3}+3^{2/3} \quad\mbox{(here we used: }\ 2x^61=5, \ \ (x^62)x^2=3^{1/3}). $$ So the denominator becomes $53^{1/3}+3^{2/3}$. 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 $$ ((2n1)(n2)x^2+x^4) \ \times \ ((5n1)+(2n2)x^2+(n5)x^4) $$ $$ = (n  1) (n^2  18 n + 1) $$ is an integer.
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Updated on December 01, 2022Comments

Admin 11 months
Rationalize the denominator of this fraction
$$\frac{1}{1+\sqrt{12}\sqrt[3]{3}\sqrt[3]{9}}$$

2012ssohn almost 7 yearsWelcome to StackExchange! Please let us know what you've tried so we can help you better!


Admin almost 7 yearsAnd 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 $\displaystyle \huge \ddot \smile$.

Alex almost 7 yearsHappy New Year everyone! If you have an alternative solution, please work on getting this question reopened. (btw "This should be a comment" now makes no sense at all.)

Admin almost 7 yearsThis question can't be reopened because the op did not provide his work. And even if he provides it now it is unlikely that this question will reopen.