Simplifying Radicals

If we prime factor 200 we get: 2(2)(2)(5)(5).

We have a pair of 2s and a pair of 5s that come outside the radical.

The other 2 is left over.

We get $$2\left(5\right)\ \sqrt[]{2}$$ .

Simplified we get $$10\ \sqrt[]{2}$$

If we check to see if this is a perfect cube, we see that it is!

The answer is simply 8 because 8(8)(8) = 512

First we work with the 8. If we prime factor, we get 2(2)(2). That is one pair of 2's and one left over giving us $$2\ \sqrt[]{2}$$ .

Now for the variables we can either write them out and see how many pairs there are, or we can recall pairs are 2, and divide the exponent (that is how many of the variable we have) by 2 to see how many pairs we have:

$$\sqrt[]{x^4}=x^{\frac{4}{2}}=x^2$$

$$\sqrt[]{y^7}=y^{\frac{7}{2}}=y^{3.5}=y^3\ \sqrt[]{y}$$

Putting it together we get:

$$2x^2y^3\ \sqrt[]{2y}$$

We look under the radical at the radicand.

Is the number prime? If so, the number is all good we can't simplify. If not, we need to prime factor to see if the number can be simplified (any pairs of factors).

We look at the radicand and if any variable has an exponent > 1, then it is not completely simplified.

$$\sqrt[]{21}$$ is prime, so it is completely simplified.

$$3\ \sqrt[]{5}$$ , the 5 is prime, so it is completely simplified.

$$3x^5y^{9\ }\sqrt[]{10xy}$$ . 10 prime factored is 2(5), so it can't be simplified. Each variable in the radicand has an exponent of 1, so we are good there. This is completely simplified.

In the other two, 8 is prime factored to 2(2)(2). There is a pair of 2's so it isn't simplified.

The 36 is a perfect square, so it can be simplified and the variables all have exponents greater than 2.

If we check 81 isn't a perfect cube.

When we prime factor get 3(3)(3)(3).

That one set of three 3's with one left over.

$$3\ \sqrt[3]{3}$$

320 isn't a perfect square.

We prime factor and we get: 2(2)(2)(2)(2)(2)(5)

That would be $$2\cdot2\cdot2\ \sqrt[]{5}$$

That is $$8\ \sqrt[]{5}$$

We check and 320 isn't a perfect cube.

When we prime factor we get:

2(2)(2)(2)(2)(2)(5)

that is 2 groups of 3 twos.

$$2\cdot2\ \sqrt[3]{5}$$ which is $$4\ \sqrt[3]{5}$$