Question

Rationalizing the denominator?

Answer 1

$$\frac{4 + \sqrt[3]{3}}{\sqrt[3]{6}}$$

Answer 2

\[\huge\color{blue}{ \frac{(4+\sqrt[3]{6}) \times \sqrt[3]{6}}{\sqrt[3]{6} \times \sqrt[3]{6}} }\]

Answer 3

The denominator would be:
$$3 \sqrt[3]{2}$$
How do we we simplify further than that?

Answer 4

No, it wouldn't be that.
\[\huge\color{blue}{ \frac{4 \sqrt[3]{6}+6}{6} }.\]\[\huge\color{red}{ B/c~~both~~\sqrt[3]{6} \times \sqrt[3]{6}~~ became~6 }.\]

Answer 5

Now divide top and bottom by 2.

Answer 6

What property makes:
$$\sqrt[3]{6} \times \sqrt[3]{6} = 6 ?$$

Answer 7

My bad.\[\huge\color{blue}{ \frac{4+ \sqrt[3]{6}}{ \sqrt[3]{6} } }\]\[\huge\color{blue}{ \frac{(4+ \sqrt[3]{6}) \times \sqrt[3]{36}}{ \sqrt[3]{6}\times \sqrt[3]{36} } }\]

Answer 8

because now it's going to be.\[\sqrt[3]{6^3}~~~~~~~~->~~~~~~~~6\]

Answer 9

How did you get \(\sqrt[3]{36}\) ?
I thought we multiply by the conjugate?

Answer 10

I multiplied top and bottom times \[\sqrt[3]{36}\]

Answer 11

Okay, so where did you get:
\[\sqrt[3]{6^3}\]

Answer 12

lets expand the parenthesis on top and bottom.
\[\huge\color{blue}{ \frac{4\sqrt[3]{36}+6}{6} }~~~->~~~\huge\color{blue}{ \frac{2\sqrt[3]{36}+3}{3} }\]
the reason I said \[\sqrt[3]{6^3}\] is because when I simplify the bottom, \[\sqrt[3]{6} \times \sqrt[3]{36}~~->~~\sqrt[3]{6} \times \sqrt[3]{6^2}~~->~~\sqrt[3]{6 \times 6^2}~~~->~~~\sqrt[3]{6^3}~~~->~~~6.\]

Answer 13

Thank you :-).

Answer 14

YW!