Simplify the Nested Cube Root: ∛(∛(512x^27))

Nested Radical Simplification with Perfect Powers

Complete the following exercise:

512x2733= \sqrt[3]{\sqrt[3]{512x^{27}}}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:10 Let's solve this math problem together.
00:13 Imagine we have a number A, raised to the power of B, within a root of order C.
00:19 The answer will be A to the root of B times C.
00:23 Let's use this rule in our example. We'll calculate the order of the products.
00:36 When we take the root of a product, like A times B,
00:40 we can express it as the product of each term's root.
00:45 Let's apply this to our exercise and simplify the roots.
00:54 We'll break down 512 into two to the power of nine.
00:59 When A is raised to the power B in root C,
01:03 it becomes A to the power of B divided by C.
01:07 We'll use this formula to find the power quotients for our question.
01:18 And that's how we solve this problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Complete the following exercise:

512x2733= \sqrt[3]{\sqrt[3]{512x^{27}}}=

2

Step-by-step solution

To solve the given problem, we'll follow these steps:

  • Step 1: Simplify the innermost cube root 512x273\sqrt[3]{512x^{27}}.
  • Step 2: Simplify the next cube root ()3\sqrt[3]{(\cdot)} from the result of step 1.

Let's go through each step:

Step 1: Consider the expression 512x273\sqrt[3]{512x^{27}}.
First, evaluate 5123\sqrt[3]{512}. Since 512=83512 = 8^3, we have 5123=8\sqrt[3]{512} = 8.
For x273\sqrt[3]{x^{27}}, use the property amn=am/n\sqrt[n]{a^m} = a^{m/n}:
x273=x27/3=x9\sqrt[3]{x^{27}} = x^{27/3} = x^9.
Thus, 512x273=8x9\sqrt[3]{512x^{27}} = 8x^9.

Step 2: Now, evaluate the outer cube root 8x93\sqrt[3]{8x^9}.
83=2\sqrt[3]{8} = 2 since 8=238 = 2^3.
For x93\sqrt[3]{x^9}, again use the rule amn=am/n\sqrt[n]{a^m} = a^{m/n}:
x93=x9/3=x3\sqrt[3]{x^9} = x^{9/3} = x^3.
Therefore, 8x93=2x3\sqrt[3]{8x^9} = 2x^3.

In conclusion, the simplified expression is 2x32x^3.

Thus, the solution to the problem is 2x3 2x^3 , which corresponds to choice 3.

3

Final Answer

2x3 2x^3

Key Points to Remember

Essential concepts to master this topic
  • Radical Rule: Work from innermost radical outward, step by step
  • Technique: x273=x27/3=x9 \sqrt[3]{x^{27}} = x^{27/3} = x^9 using power division
  • Check: (2x3)3=8x9 (2x^3)^3 = 8x^9 should equal inner result ✓

Common Mistakes

Avoid these frequent errors
  • Trying to simplify both radicals simultaneously
    Don't attempt 512x2733=512x279 \sqrt[3]{\sqrt[3]{512x^{27}}} = \sqrt[9]{512x^{27}} = wrong exponent rules! This ignores proper order of operations. Always work from the innermost radical first, then apply the next radical to that result.

Practice Quiz

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Solve the following exercise:

\( \sqrt{\sqrt{4}}= \)

FAQ

Everything you need to know about this question

Why can't I just combine the cube roots into one radical?

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Order matters! Nested radicals mean you apply one operation, then another. Think of it like functions - you can't skip steps. 512x2733 \sqrt[3]{\sqrt[3]{512x^{27}}} is different from 512x276 \sqrt[6]{512x^{27}} .

How do I know when a number is a perfect cube?

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Look for numbers that equal something cubed. For example: 8=23 8 = 2^3 , 27=33 27 = 3^3 , 512=83 512 = 8^3 . For variables, the exponent should be divisible by 3.

What if the exponent isn't divisible by 3?

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You can still simplify! Use xn3=xn/3 \sqrt[3]{x^n} = x^{n/3} even with fractions. For example: x103=x10/3=x313 \sqrt[3]{x^{10}} = x^{10/3} = x^{3\frac{1}{3}} .

How can I check if my final answer is correct?

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Work backwards! If your answer is 2x3 2x^3 , then cube it twice: (2x3)3=8x9 (2x^3)^3 = 8x^9 , then (8x9)3=512x27 (8x^9)^3 = 512x^{27} . This should match your original expression inside!

Is there a shortcut for nested radicals?

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Not really - and shortcuts often lead to errors! The step-by-step approach is actually faster because you avoid mistakes. Plus, it helps you understand what's happening mathematically.

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