Mathematics, 03.08.2020 22:01, zachstonemoreau
Simplify $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$
Answers: 2
![x=\sqrt[3]{3}](/tpl/images/0717/0057/74cb0.png)
and ![x^2=\sqrt[3]{9}](/tpl/images/0717/0057/2df3b.png)
. Then![\dfrac{2\sqrt[3]{9}}{1+\sqrt[3]{3}+\sqrt[3]{9}}=\dfrac{2x^2}{1+x+x^2}](/tpl/images/0717/0057/cbe88.png)
. The motivation for this is the rule for factoring a difference of cubes:
![\dfrac{2\sqrt[3]{9}}{1+\sqrt[3]{3}+\sqrt[3]{9}}=\dfrac{2\sqrt[3]{9}(1-\sqrt[3]{3})}{1-3}=\sqrt[3]{9}(\sqrt[3]{3}-1)=3-\sqrt[3]{9}](/tpl/images/0717/0057/8b7cc.png)
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Simplify $\frac{2\sqrt[3]9}{1 + \sqrt[3]3 + \sqrt[3]9}.$...
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