Solve the Nested Radical: Cube Root of Square Root of 64

Q: Why can't I just work from the inside out like normal?

You can work inside-out for simple cases like this one! $ \sqrt{64} = 8 $, then $ \sqrt[3]{8} = 2 $. But using exponential form is more reliable for complex nested radicals and teaches you the underlying pattern.

Q: How do I remember the exponent rule for nested radicals?

Think of it as "multiply the exponents": $ \sqrt[3]{\sqrt{x}} = x^{1/2 \cdot 1/3} = x^{1/6} $. The outer exponent gets multiplied by the inner exponent!

Q: What does 64^(1/6) actually mean?

$ 64^{1/6} $ means "what number raised to the 6th power equals 64?" Since $ 2^6 = 64 $, the answer is 2. It's the 6th root of 64!

Q: Can I use this method with any nested radical?

Yes! Convert each radical to exponential form: $ \sqrt[n]{x} = x^{1/n} $. Then use $ (a^m)^n = a^{mn} $ to multiply exponents. This works for any level of nesting!

Q: Why is 64 written as 2^6?

Recognizing perfect powers makes calculations easier! $ 64 = 2^6 $ because $ 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64 $. This lets us simplify $ 64^{1/6} = (2^6)^{1/6} = 2^1 = 2 $ easily.

Nested Radicals with Exponent Properties

Solve the following exercise:

$\sqrt[3]{\sqrt{64}}=$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve the following problem

00:03 A "regular" root is of the order 2

00:09 When we have a number (A) in a root of the order (B) in a root of the order (C)

00:14 The result equals the number (A) in a root of the order of their product (B times C)

00:18 We will apply this formula to our exercise

00:23 Let's calculate the order of multiplication

00:30 When we have a number (A) to the power of (B) in a root of the order (C)

00:36 The result equals the number (A) to the power of their quotient (B divided by C)

00:39 We will apply this formula to our exercise

00:46 This is the solution

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Solve the following exercise:

$\sqrt[3]{\sqrt{64}}=$

Step-by-step solution

To solve this problem, we'll express the nested roots using exponents and then simplify:

We start with the inner expression:

$\sqrt{64}$ is equivalent to $64^{1/2}$ .

Next, apply the cube root:

$\sqrt[3]{64^{1/2}}$ is equivalent to $(64^{1/2})^{1/3}$ .

Using properties of exponents, we simplify the expression:

$(64^{1/2})^{1/3} = 64^{(1/2) \cdot (1/3)} = 64^{1/6}$ .

Now, evaluate $64^{1/6}$ :

Since $64 = 2^6$ , we have:

$64^{1/6} = (2^6)^{1/6} = 2^{6 \cdot (1/6)} = 2^{1} = 2$ .

Therefore, the solution to the exercise $\sqrt[3]{\sqrt{64}}$ is $\mathbf{2}$ .

Final Answer

Key Points to Remember

Essential concepts to master this topic

Rule: Convert nested radicals to exponential form before simplifying
Technique: $\sqrt[3]{\sqrt{64}} = (64^{1/2})^{1/3} = 64^{1/6}$
Check: Verify $2^6 = 64$ , so $64^{1/6} = 2$ ✓

Common Mistakes

Avoid these frequent errors

Evaluating radicals separately without using exponent laws
Don't calculate $\sqrt{64} = 8$ then $\sqrt[3]{8} = 2$ = correct but inefficient! This approach works here but fails with complex nested radicals. Always convert to exponential form using $(a^m)^n = a^{mn}$ for systematic solutions.

Practice Quiz

Test your knowledge with interactive questions

Solve the following exercise:

$\sqrt[10]{\sqrt[10]{1}}=$

FAQ

Everything you need to know about this question

Why can't I just work from the inside out like normal?

You can work inside-out for simple cases like this one! $\sqrt{64} = 8$ , then $\sqrt[3]{8} = 2$ . But using exponential form is more reliable for complex nested radicals and teaches you the underlying pattern.

How do I remember the exponent rule for nested radicals?

Think of it as "multiply the exponents": $\sqrt[3]{\sqrt{x}} = x^{1/2 \cdot 1/3} = x^{1/6}$ . The outer exponent gets multiplied by the inner exponent!

What does 64^(1/6) actually mean?

$64^{1/6}$ means "what number raised to the 6th power equals 64?" Since $2^6 = 64$ , the answer is 2. It's the 6th root of 64!

Can I use this method with any nested radical?

Yes! Convert each radical to exponential form: $\sqrt[n]{x} = x^{1/n}$ . Then use $(a^m)^n = a^{mn}$ to multiply exponents. This works for any level of nesting!

Why is 64 written as 2^6?

Recognizing perfect powers makes calculations easier! $64 = 2^6$ because $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$ . This lets us simplify $64^{1/6} = (2^6)^{1/6} = 2^1 = 2$ easily.

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