calc

[Alex]

i have x ^ 1/2 - x ^ 1/3 (the square root of x minus the cube root of x)

can i factor it to something like this: x( 1 ^ 1/2 - 1 ^ 1/3) ? x(square root of 1 minus cube root of 1) ?

if not how would i go about factoring that expression?

thx!

 

[GiLtY]

Don't think so, 1 to the any power is 1, so your factor would give you zero.

--GiLtY

 

[phreakah]

would x^1/3(x^1/6 - 1) work?

 

[GiLtY]

Originally posted by: phreakah
would x^1/3(x^1/6 - 1) work?

Yap

--GiLtY

 

[Alex]

nice thx eh!

 

[Alex]

Originally posted by: phreakah
would x^1/3(x^1/6 - 1) work?

on second thought, i dont think that would work at all


edit: my solution is:

x ^ 1/2 ( 1 - x ^ 2/3)

which would give x ^ 1/2 - x ^ 2 / 6 = x ^ 1/3

 

[phreakah]

i just used the factor feature of my ti89..

 

[eakers]

Originally posted by: franguinho

Originally posted by: phreakah
would x^1/3(x^1/6 - 1) work?

on second thought, i dont think that would work at all


edit: my solution is:

x ^ 1/2 ( 1 - x ^ 2/3)

which would give x ^ 1/2 - x ^ 2 / 6 = x ^ 1/3

you have to add exponents of the same base when multiplying.

 

[Alex]

Originally posted by: eakers

Originally posted by: franguinho

Originally posted by: phreakah
would x^1/3(x^1/6 - 1) work?

on second thought, i dont think that would work at all


edit: my solution is:

x ^ 1/2 ( 1 - x ^ 2/3)

which would give x ^ 1/2 - x ^ 2 / 6 = x ^ 1/3

you have to add exponents of the same base when multiplying.

so is my 'solution' garbage?

 

[GiLtY]

Originally posted by: franguinho

Originally posted by: eakers

Originally posted by: franguinho

Originally posted by: phreakah
would x^1/3(x^1/6 - 1) work?

on second thought, i dont think that would work at all


edit: my solution is:

x ^ 1/2 ( 1 - x ^ 2/3)

which would give x ^ 1/2 - x ^ 2 / 6 = x ^ 1/3

you have to add exponents of the same base when multiplying.

so is my 'solution' garbage?

What math are you in...

--GiLtY

 

[Alex]

k k i got it i was totally wrong my bad !!!

thx for the help guys