Question

2/x^1/3

Answer 1

\[\frac{ 2 }{ x^{1/3} }\]What is your question?

Answer 2

how to solve it ?

Answer 3

It's the simplest form. What else can we do?

Answer 4

I'm sorry, rationalize the denominator: 2/x^1/3

Answer 5

because there is cuberoot in denominator so you need rationalizing the denominator

do you know it how ?

Answer 6

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

Answer 7

\[\frac{ 2 }{ \sqrt[3]{x} } \times \frac{ \sqrt[3]{x^2} }{ \sqrt[3]{x^2} }\]

Answer 8

why did you put the x to the power of two?

Answer 9

Because if you multiply the denominators together, you will get \[\sqrt[3]{x^3} = x\]

Answer 10

Thank you so much! @micahwood50

Answer 11

@micahwood50 so this not is right

Answer 12

when you need rationalzing the denominator so than just multiplie the numerator and denominator with denominator

ok ?

so in this case with cuberoot x
and hence will get in numerator 2cuberootx and in denominator x

ok ?

Answer 13

do you understand it now ?

Answer 14

so for example when there is

1/sqrtx so than need multiplie the numerator and denominator by sqrt x
and will get sqrt x /x

ok ?

Answer 15

Yeah? That's the point... And now do the numerators and then you will get \[\frac{ 2\sqrt[3]{x^2} }{ x }\]

Answer 16

not is right because in numerator inside cuberoot will be just x and not x squared

Answer 17

so because the denominator is cuberoot x and not x squared

Answer 18

\[\frac{ 2 }{ \sqrt[3]{x} } \times \frac{ \sqrt[3]{x^2} }{ \sqrt[3]{x^2} } = \frac{ 2\sqrt[3]{x^2} }{ x }\]Well, Didn't you noticed that I basically multiply it by one, since \[\frac{ \sqrt[3]{x^2} }{ \sqrt[3]{x^2} } = 1\] So it doesn't affect the term.

I just want to "remove" the root in denominator.\[\sqrt[3]{x}\sqrt[3]{x^2} = x^{1/3}x^{2/3} = x^{3/3} = x\]

I argue that my answer is still right.