Question

How does this equal that(problem inside)?

Answer 1

\[\frac{ \sqrt[3]{5} }{ \sqrt[3]{st^2} } = \frac{ \sqrt[3]{5s^2t} }{ st }\]

My Steps:
\[\frac{ \sqrt[3]{5} }{ \sqrt[3]{st^2} } * \frac{ \sqrt[3]{st^2} }{ \sqrt[3]{st^2} } * \frac{ \sqrt[3]{st^2} }{ \sqrt[3]{st^2} } = \frac{ \sqrt[3]{5s^2t^6} }{ \sqrt[3]{s^3t^6} }\]

I don't know where I went wrong

Answer 2

\[\Large \frac{ \sqrt[3]{5} }{ \sqrt[3]{st^2} } * \frac{ \sqrt[3]{st^2} }{ \sqrt[3]{st^2} } * \frac{ \sqrt[3]{st^2} }{ \sqrt[3]{st^2} } = \frac{ \sqrt[3]{5s^2t^6} }{ \sqrt[3]{s^3t^6} }\]your font was too small on my screen.

Answer 3

You did too much work, for one. Make it easier like so:
\[\Large \frac{ \sqrt[3]{5} }{ \sqrt[3]{st^2} }*\frac{ \sqrt[3]{s^2 t} }{ \sqrt[3]{s^2 t} }\]

all you need to do is get the exponents under the bottom cube root to be 3's

Answer 4

Where did you get this from? \[\frac{ \sqrt[3]{s^2t} }{ \sqrt[3]{s^2t} }\]

Answer 5

@agent0smith in order to rationalize the denominator, you need to multiply by two copies of the denominator

Answer 6

No you don't...

Answer 7

just get photomath it works for everything but graphs and word problems https://www.photomath.net/en/

Answer 8

You want the exponents under that cube root to all become 3's, so figure out how many to put on each

Eg. if you had \[\Large \frac{ 5 }{ \sqrt[100]{x}}\]
then you do this \[\Large \frac{ 5 }{ \sqrt[100]{x}}*\frac{ \sqrt[100]{x^{99}} }{ \sqrt[100]{ x^{99} }}\]

Answer 9

Example:

\[\Large \frac{1}{\sqrt[3]{x}} = \frac{1}{\sqrt[3]{x}}*{\color{red}{\frac{\sqrt[3]{x}}{\sqrt[3]{x}}}}*{\color{red}{\frac{\sqrt[3]{x}}{\sqrt[3]{x}}}}\]

\[\Large \frac{1}{\sqrt[3]{x}} = \frac{\sqrt[3]{x}*\sqrt[3]{x}}{\sqrt[3]{x}*\sqrt[3]{x}*\sqrt[3]{x}}\]

\[\Large \frac{1}{\sqrt[3]{x}} = \frac{\sqrt[3]{x*x}}{\sqrt[3]{x*x*x}}\]

\[\Large \frac{1}{\sqrt[3]{x}} = \frac{\sqrt[3]{x^2}}{\sqrt[3]{x^3}}\]

\[\Large \frac{1}{\sqrt[3]{x}} = \frac{\sqrt[3]{x^2}}{x}\]

Answer 10

@jim_thompson5910 do you want to try that with my previous example...?

No, you multiply by enough exponents that add up to the root exactly.

Answer 11

true it does work, so that's a valid alternative

Answer 12

hey @agent0smith can you hlp me with a WORD PROBLEM

Answer 13

I learned it this way from an algebra 2 book, teaching it a couple of years ago

in general:\[\Large \frac{ 1 }{ \sqrt[a]{x^b} }*\frac{ \sqrt[a]{x^{a-b}} }{ \sqrt[a]{x^{a-b}} }\]

Answer 14

@-Welp- the way you did it is fine too, but that last exponent should be a 4, i fixed it \[\Large \frac{ \sqrt[3]{5} }{ \sqrt[3]{st^2} } * \frac{ \sqrt[3]{st^2} }{ \sqrt[3]{st^2} } * \frac{ \sqrt[3]{st^2} }{ \sqrt[3]{st^2} } = \frac{ \sqrt[3]{5s^2t^4} }{ \sqrt[3]{s^3t^6} }\]

Answer 15

then just simplify a little\[\Large \frac{ \sqrt[3]{5s^2t^4} }{ \sqrt[3]{s^3t^6} } = \frac{ t\sqrt[3]{5s^2t} }{ st^2 } \]