You're comfortable with square roots, like how $\sqrt{16}$ is 4. Now, let's generalize this concept. Instead of just finding a number that multiplies by itself twice to get another, we can look for numbers that multiply by themselves three, four, or any number of times. This is the idea behind nth roots.

For example, the cube root of 8 is 2, because $2 \cdot 2 \cdot 2 = 8$. We write this as $\sqrt[3]{8} = 2$. The fourth root of 81 is 3, because $3 \cdot 3 \cdot 3 \cdot 3 = 81$. We write this as $\sqrt[4]{81} = 3$. The same rules you used for simplifying square roots, the product and quotient properties, also apply to these higher-order roots.

The product and quotient rules are your main tools for breaking down complex radicals into simpler forms. They work exactly the same way, no matter the index.

Let's simplify $\sqrt[3]{128}$. We're looking for the largest perfect cube that divides 128. A quick check shows us that $4^3 = 64$, and $128 = 64 \cdot 2$.

Using the product rule: $\sqrt[3]{128} = \sqrt[3]{64 \cdot 2} = \sqrt[3]{64} \cdot \sqrt[3]{2}$

Since we know $\sqrt[3]{64} = 4$, the simplified form is $4\sqrt[3]{2}$.

Now let's try simplifying $\sqrt[4]{\frac{162}{2}}$. First, simplify the fraction inside the radical: $\frac{162}{2} = 81$. So we have $\sqrt[4]{81}$.

We're looking for a number that, when multiplied by itself four times, equals 81. That number is 3, because $3^4 = 81$. So, the answer is 3.

The same logic applies to variables with exponents. To simplify a variable under an nth root, you divide its exponent by the index. The whole number part of the result is the exponent of the variable that comes outside the radical, and the remainder is the exponent of the variable that stays inside.

Let's simplify $\sqrt[3]{x^8}$.

We divide the exponent (8) by the index (3). 8 divided by 3 is 2 with a remainder of 2.

This means we can pull out $x^2$, and we are left with $x^2$ inside. The final simplified form is $x^2\sqrt[3]{x^2}$.

Now, let's put it all together with numbers and multiple variables. Consider $\sqrt[4]{80x^9y^{12}}$.

1. Factor the number: The largest perfect fourth power that divides 80 is 16 ($2^4 = 16$). So, $80 = 16 \cdot 5$.
2. Handle the x variable: Divide the exponent 9 by the index 4. This gives 2 with a remainder of 1. So we pull out $x^2$ and leave $x^1$ inside.
3. Handle the y variable: Divide the exponent 12 by the index 4. This gives 3 with a remainder of 0. So we pull out $y^3$ and leave nothing inside.

Combining these pieces, we get: $\sqrt[4]{16 \cdot 5 \cdot x^8 \cdot x^1 \cdot y^{12}} = \sqrt[4]{16} \cdot \sqrt[4]{x^8} \cdot \sqrt[4]{y^{12}} \cdot \sqrt[4]{5x} = 2x^2y^3\sqrt[4]{5x}$

There's one final, important detail. When you take an even root (square root, fourth root, etc.) of a variable expression and the result has an odd exponent, you must use absolute value bars. This ensures the result is always non-negative, because the original expression under the even root could not have been negative.

Consider $\sqrt{x^2}$. You might think it's just $x$. But what if $x = -3$? Then $\sqrt{(-3)^2} = \sqrt{9} = 3$. The result is not -3, it's $|-3|$. So, $\sqrt{x^2} = |x|$.

Let's simplify $\sqrt[4]{x^{12}y^{14}}$.

• For $x$: $12 \div 4 = 3$. We pull out $x^3$. Since the index (4) is even and the resulting exponent (3) is odd, we need absolute value: $|x^3|$.
• For $y$: $14 \div 4 = 3$ with a remainder of 2. We pull out $y^3$ and leave $y^2$ inside. The index is even and the resulting exponent is odd, so we need absolute value here too: $|y^3|$.

The simplified expression is $|x^3y^3|\sqrt[4]{y^2}$.

You only need to worry about this rule when the index is even. For odd roots, like a cube root, you never use absolute value bars because you can take an odd root of a negative number (e.g., $\sqrt[3]{-8} = -2$).

Now that you've seen how to simplify nth roots, test your skills.