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Algebra/Algebra 2 Math Question


Math Problem
Math Problem  
Good afternoon,

I solved the problem below and arrived at the wrong answer. I cannot figure out what I did incorrectly ans was hoping you could help with it! I have also attached an image of what it looks like formatted correctly.

Thank you for your time :)

Simplify. Assume that all variables are positive.

The answer I got is b√6b/6

To do this, I would first put all terms under the cube root.
Note that b^2 is √b^6.
This makes the fraction into √[b^6/(6b^2)].
Since b^6/b^2 b^4, that is √(b^4/6).

Since we have a b^4 and are taking a cube root, that is the same as b√b.
This makes the fraction into b√(b/6).

Now since we don't like cube roots in the denominator of a fraction, multiply the top and bottom by 6^2, which is 36.

That gives b√(36b/216) = (b/6)√(36b).

It looks like what was forgotten in finding the solution is that the 6 has a cube root as well.
This makes 1/√6 =  √36 / 6.


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Scott A Wilson


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