How do you minus a cubed root from a square root?

Question

I have a question on my hw that wants me to simplify the equation: the cubed root of 6 minus the square root of 6. Can this be done, and if so how?

Answer 1

It can't; at least, not in the way you might think. It's like subtracting y from x; two different things that just don't mesh.

It can, however, simplify in this way.

cuberoot(6) - sqrt(6)

Let's change both of these to fractional exponents.

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

Now, let's pull out the number with the smaller power.

6^(1/3) [1 - 6^(1/2 - 1/3)]
6^(1/3) [1 - 6^(3/6 - 2/6)]
6^(1/3) [1 - 6^(1/6)]

Although this is a different form of the expression, it did nothing to simplify things in the way you might wish.

You can, however, multiply these and it becomes simplified.

6^(1/3) * 6^(1/2) has the same base but different exponent, so this would become

6^(1/3 + 1/2) = 6^(2/6 + 3/6) = 6^(5/6)

So as for simplifying, multiplying/dividing, yes; adding and subtracting, no.

Answer 2

You can't minus a cubed root from a square root and I can't remember how to simplify. Hope this helps ^^; Yeah the one above me, that is Trig./ Alg.3 way to go. ;-)