Simplifying Cube Roots Please Help!!!

[Mickeyy21]

I don't even know where to start wight the cube roots.

Cube Root:

^3√(108(x^15)(y^29)) or (108(x^15)(y^29))^1/3 I know the answer is( 3(x^5)(y^9)√y^2), but I have no idea how to get there.

 

[stapel]

I don't even know where to start wight the cube roots.

Cube Root:

^3√(108(x^15)(y^29)) or (108(x^15)(y^29))^1/3 I know the answer is( 3(x^5)(y^9)√y^2), but I have no idea how to get there.

Would you know how to get started if this were a square root? If so, then use the same methods (factor, pull out anything you've got two factors of, etc). If not, then please specify so. Thank you!

 

[Bob Brown MSEE]

Work backwards

When you are given the answer it is often useful to work backward.
In this case, cube the answer and see if you get some insights.

 

[Quaid]

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Bob, what's that equation anyway?

There's no equals sign, Denis.

Are you asking about the ^3√ notation for cube root?

Same as (108(x^15)(y^29))^(1/3)
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[Deleted member 4993]

ֺ There's no equals sign, Denis.ֺ

Normally that requires a penalty in the corner. But then "ice-hockey" players loose eye-sight in the glare of ice - so we will skip that this time.

By the way, "field-hockey" players or cricket players - do not suffer such maladies. They play the game on green grass - soothing to the eye....

 

[Quaid]

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or (108(x^15)(y^29))^(1/3)

It's not a cube root, unless you put grouping symbols (shown in red above) around the exponent (remember: Order of Operations).

the answer is

(3(x^5)(y^9)√y^2)

Did you mistype the factor highlighted in red above? Is that notation supposed to denote y^(2/3) ?

Regardless, that answer is not correct.

We can verify that it's wrong by cubing it.

(3(x^5)(y^9)y^(2/3))^3 = 27 x^15 y^29

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