how would you simplify a cubed root radical of 25 times the radical of 125

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OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

ok, so how are 25 and 125 related?

OpenStudy (anonymous):

Both divisible by 5 and 25?

OpenStudy (anonymous):

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

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OpenStudy (anonymous):

5^3, sorry

OpenStudy (anonymous):

oh okay yeah

OpenStudy (anonymous):

\[\sqrt[3]{25} \times \sqrt{125} = 25^{1/3} \times 125^{1/2} = 5^{2/3} \times 5^{3/2} = 5^{4/6} \times 5^{9/6} = 5^{13/6}\]
Feel free to correct me, this is my first time using this equation maker.

OpenStudy (anonymous):

omg it's too long for the page HAHA

OpenStudy (anonymous):

what is the equation maker?

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OpenStudy (anonymous):

all right

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yeah, that's how to do it easily hrwhyhry

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohhhhh oka

OpenStudy (anonymous):

y
mn

OpenStudy (anonymous):

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

OpenStudy (anonymous):

okay thank you so much!

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OpenStudy (anonymous):

so, you got it?

OpenStudy (anonymous):

i think soooo

OpenStudy (anonymous):

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

OpenStudy (anonymous):

sorry its supposed to be the x- the 3 one

OpenStudy (anonymous):

????

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OpenStudy (anonymous):

just a second

OpenStudy (anonymous):

thank you :)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

is that the problem? is it equal to something?

OpenStudy (anonymous):

no thats the problem. You only have to simplify it

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OpenStudy (anonymous):

ok

OpenStudy (anonymous):

so what is sqrt(12)?

OpenStudy (anonymous):

what are factors of 12?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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OpenStudy (anonymous):

so you can separate x and -3^(1/3) because they have the same denominator and they are part of a subtraction operation

OpenStudy (anonymous):

ok...

OpenStudy (anonymous):

thank u for the help by the way

OpenStudy (anonymous):

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

OpenStudy (anonymous):

np

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OpenStudy (anonymous):

can you simplify that? focus on the 3's

OpenStudy (anonymous):

u can divide 3 by 3 and get one right

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so 3^(1/3) / 3^(1/2) is 3^(1/3) * 3^-(1/2)...can you can add these exponents --> 1/3 = 2/6, 1/2 = 3/6, need to have similar fractions....so 2/6 - 3/6 = -1/6

OpenStudy (anonymous):

so you have 3^(-1/6)

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OpenStudy (anonymous):

which is 1/(3^1/6)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yeah, like in the last message

OpenStudy (anonymous):

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

OpenStudy (anonymous):

wait, x-1

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