Solve Nested Radicals: Sixth Root of √64 × Fourth Root of ∛16

Question

Complete the following exercise:

6461634= \sqrt[6]{\sqrt{64}}\cdot\sqrt[4]{\sqrt[3]{16}}=

Video Solution

Solution Steps

00:00 Solve the following problem
00:03 A 'regular' root raised to the second power
00:11 When there is a root of order (C) to root (B)
00:15 The result equals the root of the product of the orders
00:19 Apply this formula to our exercise
00:29 When we have a root of order (C) on number (A) to the power of (B)
00:32 If we have a product (A times B) in a root of order (C)
00:36 We can divide the product into two roots of order (C)
00:40 Apply this formula to our exercise, in reverse
00:49 Let's calculate the product
00:53 This is the solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Simplify 64\sqrt{64}.
  • Step 2: Simplify with 6\sqrt[6]{}.
  • Step 3: Simplify 163\sqrt[3]{16}.
  • Step 4: Simplify with 4\sqrt[4]{}.
  • Step 5: Combine the simplified results.

Now, let's work through each step:

Step 1: Simplify 64\sqrt{64}.
Since 64=8264 = 8^2, we have 64=8\sqrt{64} = 8.

Step 2: Simplify 86\sqrt[6]{8} using exponent rules:
86=81/6\sqrt[6]{8} = 8^{1/6}.

Step 3: Simplify 163\sqrt[3]{16}.
Since 16=2416 = 2^4, then 163=(24)1/3=24/3\sqrt[3]{16} = (2^4)^{1/3} = 2^{4/3}.

Step 4: Simplify 24/34\sqrt[4]{2^{4/3}} using exponent rules:
We have 24/34=(24/3)1/4=24/12=21/3\sqrt[4]{2^{4/3}} = (2^{4/3})^{1/4} = 2^{4/12} = 2^{1/3}.

Step 5: Combine simplified results:
We have 81/621/3=(23)1/621/3=23/621/3=21/221/3=2(1/2+1/3)=2(3/6+2/6)=25/6.8^{1/6} \cdot 2^{1/3} = (2^3)^{1/6} \cdot 2^{1/3} = 2^{3/6} \cdot 2^{1/3} = 2^{1/2} \cdot 2^{1/3} = 2^{(1/2 + 1/3)} = 2^{(3/6 + 2/6)} = 2^{5/6}.

Convert to a common root form:
The result is equivalent to 25/6=21012=1024122^{5/6} = \sqrt[12]{2^{10}} = \sqrt[12]{1024}.

Therefore, the solution to the problem is 102412 \sqrt[12]{1024} .

Answer

102412 \sqrt[12]{1024}