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ok, you have 25^(1/3) * 125^(1/2)?

yes

ok, so how are 25 and 125 related?

Both divisible by 5 and 25?

ok, cool, so 25 = 5^2 and 125 = 5^2

5^3, sorry

oh okay yeah

omg it's too long for the page HAHA

what is the equation maker?

all right

the answer is 5^{13/6} sorry it's too long haha

yeah, that's how to do it easily hrwhyhry

okay.. how on earth did you get that though?

so you have 25=5^2, and 125=5^3....so you have (5^2)^(1/3) * (5^3)^(1/2)

you can then add the exponents...here: \[(5^2)^(1/3)\]

ohhhhh oka

y
mn

(5^2)^(1/3) --> the square (2) and the (1/3) and now add together, 2/3

okay thank you so much!

so, you got it?

i think soooo

now how would you do x-\[\sqrt[3]{3}\div \sqrt{12}\]

sorry its supposed to be the x- the 3 one

????

just a second

thank you :)

\[(x-\sqrt[3]{3})/\sqrt{12}\]

is that the problem? is it equal to something?

no thats the problem. You only have to simplify it

ok

so what is sqrt(12)?

what are factors of 12?

the square root of 4 times the square root of 3?

right on, so sqrt(4) is simple, sqrt(3) is good because the problem has 3^(1/3)

is the answer 1 over x^1/6 times 2?

1/(x^1/6 * 2)?

yes

\[(x-(3^{1/3}))/(2*(3^{1/2}))\]

i had that... i just dont kmow if u can get rid of the three

ok...

thank u for the help by the way

first look at 3^(1/3) / (2 * (3^1/2))

np

can you simplify that? focus on the 3's

u can divide 3 by 3 and get one right

when you divide by exponents you and subtract the exponents from values with the same base

so you have 3^(-1/6)

which is 1/(3^1/6)

yes and u have to bring it below the division sign since it is negative?>

yeah, like in the last message

is the answer x over 3^1/6 *2

wait, x-1

(x-1)/(2*(3^1/6))

\[(x-1)/(2*(3^1/6))\]

\[(x-1)/(2*(3^{1/6}))\]

how's that look?

Thank you sooo much! I think thats right.....