The Purplemath Forums

by **TehNyanCat** on Wed Jun 29, 2011 6:28 pm

Help please?

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by **little_dragon** on Wed Jun 29, 2011 7:19 pm

TehNyanCat wrote: $\mbox{True or false:}$

$\left(54x^{-3}y^9 z^7)^{(1/3)}\, =\, 3\sqrt[3]{2}x^{-1} y^3 z^{(2/3)}$

if 7 & 2 r rite, then wat do U think?

:wave:

by **TehNyanCat** on Wed Jun 29, 2011 7:44 pm

little_dragon wrote:
TehNyanCat wrote: $\mbox{True or false:}$

$\left(54x^{-3}y^9 z^7)^{(1/3)}\, =\, 3\sqrt[3]{2}x^{-1} y^3 z^{(2/3)}$

if 7 & 2 r rite, then wat do U think?

"7 and 2"? I'm a bit confused... :confused:

The furthest I can go is $\sqrt[3]{54x^{-3}y^9 z^7}$ .

by **maggiemagnet** on Thu Jun 30, 2011 12:28 pm

I think the previous poster was referring to the powers on the variable "z". How did you do the simplification in your first post? If you could show us your steps, it will be easier to explain what is going on. Thank you!

:clap:

by **TehNyanCat** on Thu Jun 30, 2011 4:11 pm

maggiemagnet wrote:I think the previous poster was referring to the powers on the variable "z". How did you do the simplification in your first post? If you could show us your steps, it will be easier to explain what is going on. Thank you!

The thing is, I didn't simplify the problem at all. It was like that originally.

by **maggiemagnet** on Thu Jun 30, 2011 5:59 pm

TehNyanCat wrote:The thing is, I didn't simplify the problem at all. It was like that originally.

Okay; then do the simplification yourself, and see if you get what they got. It would be helpful to pay particular attention to the powers on the variable "z", like the first reply suggested.

:clap:

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