When talking about simplifying radical expressions, we must necessarily remember radical numbers. In this way, when a number is not integer or fractional, we will call it radical, for example, √2; √3; ∛7. These radicals are different according to their degree, defined by the root index, that is to say that in ∛7 the radical is of the third degree and in √2 the radical is of the second degree. Similarly, there are similar terms; These have an equal radicand (number within the root), as we see in √2 and 5√2. On the contrary, √3 and 5√2 are not similar because their radicand is different.

To fully understand how to work with radical numbers arithmetically and to simplify expressions with these terms, we will carry out the following exercises. In addition, the different radical-exponent properties will be disclosed, applying these properties.

**Example 1**: Simplify the following expression **∜****(****√****x****³ ·****16****√****x) **

To start with the simplification within the fourth root, we will apply the following property:

^{n}√ ·^{n}√b = ^{n}√ (a · b) (1)

Applying the property to the expression we can rewrite it in the following way:

∜ [16 · √ (x³ · x)]

Recalling the power property at^{n}^{·}a^{m }= to^{(n + m)}:

∜ (16 · √x^{4})

Having this new expression, we will apply the following radical-exponent property:

^{n}√a^{m }= to^{m / n} (2)

By applying it, we will obtain:

∜ (16 · x²)

If we decompose the number 16, we can write it as:

16 = 2 · 2 · 2 = 2^{4}

So,

∜ (2^{4}X²)

Reapplying property (1) in the opposite way i.e. ^{n}√ (a · b) = ^{n}√ ·^{n}√ we get:

∜ (2^{4}) · ∜ (x²)

Applying property (2) to the expression:

2 x^{1/2}

If property (2) is applied in a contrary way, that is to say^{m / n} = ^{n}√a^{m }We rewrite the expression:

2 √x

The initial expression of the exercise can be simplified to 2√x, that is:

∜ (√x³ · 16√x) = 2√2

**Example 2**: Simplify the following expression ^{5}√ (32x^{5}y^{-10}z^{-35})

We will first decompose the number 32:

32 = 2 · 2 · 2 · 2 = 2^{5}

We rewrite the expression:

^{5}√ (2^{5}x^{5}y^{-10}z^{-35})

We can see that all exponents inside the root are numbers divisible by 5, that allows us to apply property (1):

^{2}√2^{5}· ^{5}√x^{5}·^{5}√y^{-10}·^{5}√z^{-35}

Applying property (2) the expression is simplified as follows:

2 · x · y^{-2}· Z^{-7}

Recalling the power property at^{-n }= 1 / a^{n}:

2 · x · 1 / y^{2}1 / z^{7 }

It could also be written as:

(2x) / (and^{2}z^{7} )

So, the simplification of the initial expression of the exercise is as follows:

^{5}√ (32x^{5}y^{-10}z^{-35}) = (2x) / (y^{2}z^{7} )