rules of exponents: simplify y^1/3(y^4/9-4y^2/9)

[Helen]

Can someone please help me with this?

Use rules of exponents to simplify the expression. Write the answer with positive exponents.

y^1/3(y^4/9-4y^2/9)

I tried this.

y^1/3(y^4/9-4y^2/9)

y2^1/3+4/9-4y^2/9

=y^1/4-4y^2/9

=y-2y^2/9

I need to know where I am going wrong. Thank you.

 

[stapel]

y^1/3(y^4/9-4y^2/9)

Does the above mean any of the following?

. . . . .y[sup:37aaq5y4]1/{3[(y^4 / (9 - 4y^2)] / 9}[/sup:37aaq5y4]

. . . . .y[sup:37aaq5y4]1/3[/sup:37aaq5y4] y[sup:37aaq5y4]4 / ([9 - 4y^2] / 9)[/sup:37aaq5y4]

. . . . .y[sup:37aaq5y4]1/3[/sup:37aaq5y4] y[sup:37aaq5y4](4 / [9 - 4y^2]) / 9[/sup:37aaq5y4]

. . . . .y[sup:37aaq5y4]1/3[/sup:37aaq5y4] y[sup:37aaq5y4]4 / [9 - 4y^2][/sup:37aaq5y4] / 9

. . . . .y[sup:37aaq5y4]1/3[/sup:37aaq5y4] y[sup:37aaq5y4]4/9[/sup:37aaq5y4] - 4y[sup:37aaq5y4]2[/sup:37aaq5y4] / 9

. . . . .y[sup:37aaq5y4]1/3[/sup:37aaq5y4] y[sup:37aaq5y4]4/9[/sup:37aaq5y4] - 4y[sup:37aaq5y4]2 / 9[/sup:37aaq5y4]

Or something else?

Thank you!

Eliz.

 

[Helen]

Eliz.
Simplify y ^ 1/3 (y ^ 4/9 - 4y ^ 2/9)
Is this any better?

 

[skeeter]

I hesitate in answering this post because your title says to "simplify". What is "simple" to one person may not be to another in this case.

If, however, you are to perform the multiplication, then ...

y[sup:1jsf3yes]1/3[/sup:1jsf3yes](y[sup:1jsf3yes]4/9[/sup:1jsf3yes] - 4y[sup:1jsf3yes]2/9[/sup:1jsf3yes])

distribute the factor y[sup:1jsf3yes]1/3[/sup:1jsf3yes] ...

y[sup:1jsf3yes]1/3 + 4/9[/sup:1jsf3yes] - 4y[sup:1jsf3yes]1/3 + 2/9[/sup:1jsf3yes]

add the exponents and you're done.

What would be "simple" to me would be rewriting the expression as ...

y[sup:1jsf3yes]5/9[/sup:1jsf3yes](y[sup:1jsf3yes]1/9[/sup:1jsf3yes] - 2)(y[sup:1jsf3yes]1/9[/sup:1jsf3yes] + 2)

 

[Helen]

Eliz. and Skeeter,
Thank you for your help. I will continue to work on this. Helen