Answer:
[tex]2\sqrt[3]{3}-\sqrt[3]{18}[/tex]
Step-by-step explanation:
We first rationalize the denominator. This means we multiply the numerator and denominator by a factor that will make the denominator a whole number.
Our denominator is [tex]\sqrt[3]{9}[/tex]; this can also be written as [tex]9^{\frac{1}{3}}[/tex].
For 9 to be a whole number, its exponent must be a whole number. We want to add 2/3 to the exponent of 9 on the bottom; this means we multiply by [tex]9^{\frac{2}{3}}[/tex]:
On the numerator, this gives us:
6(9^(2/3))-3(6^(1/3))(9^(2/3))
= [tex]6\sqrt[3]{9(9)} -3\sqrt[3]{2(3)}(\sqrt[3]{9(9)}) \\\\=6\sqrt[3]{3(3)(3)(3)}-3\sqrt[3]{2(3)(3)(3)(3)(3)}\\\\=6(3)\sqrt[3]{3}-3(3)\sqrt[3]{2(3)(3)}\\\\=18\sqrt[3]{3}-9\sqrt[3]{18}[/tex]
On the denominator, we will have:
(9^(1/3))(9^(2/3)) = 9^(3/3) = 9^1 = 9
This gives us:
[tex]\frac{18\sqrt[3]{3}-9\sqrt[3]{18}}{9}\\\\=2\sqrt[3]{3}-\sqrt[3]{18}[/tex]
