Question

\[\frac{2}{\sqrt[3]4}\] rationalize please.

Answer 1

multiple top and bottom each by
\[\sqrt[3]{16}\]

Answer 2

fiddle whenever u get time please come to my question , m not forcing

Answer 3

\[\frac{2}{\sqrt[3]{4}}*\frac{\sqrt[3]{16}}{\sqrt[3]{16}}=\frac{2*\sqrt[3]{16}}{\sqrt[3]{4^3}}=\frac{2*\sqrt[3]{2*2^3}}{4}=\frac{4*\sqrt[3]{2}}{4}\]

Answer 4

\[=\sqrt[3]{2}\]

Answer 5

so why 16? and wouldn't 2 work as well?

Answer 6

indeed, why 16 lol ?
because of the late hour maybe ...
2 would work as well - and would be simpler.

Answer 7

you multiply top and bottom by two copies of \[\large \sqrt[3]{4}\] which means you basically multiply top and bottom by \[\large (\sqrt[3]{4})^2=\sqrt[3]{4^2}=\sqrt[3]{16}\]

Answer 8

why two copies? because we need to take \[\sqrt[3]{4}\] and make it rational, so we need to multiply it by two copies of \[\sqrt[3]{4}\] to get


\[\sqrt[3]{4}\times \sqrt[3]{4}\times\sqrt[3]{4}=(\sqrt[3]{4})^3 = 4\]


Of course, what happens to the denominator also must happen to the numerator.

Answer 9

Jim,

I think what Mybrainvsme was asking was why not use 2 vs 16 - i.e.:multiply the top and bottom each by this:
\[\sqrt[3]{2}\]

\[\frac{2}{\sqrt[3]{4}}=\frac{2}{\sqrt[3]{2^2}}*\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\frac{2*\sqrt[3]{2}}{\sqrt[3]{2^3}}=\frac{2*\sqrt[3]{2}}{2}=\sqrt[3]{2}\]