Evaluate the cube root of 7 multiplied by the square root of 7 over the sixth root of 7 to the power of 5

To evaluate the expression √[3]{7} × √[2]{7} ÷ √[6]{7^5}, we can convert the roots into exponents and simplify step by step.

1. **Convert each term to exponent form:**

• Cube root of 7: 7^(1/3)
• Square root of 7: 7^(1/2)
• Sixth root of 7 to the power of 5: (7^(1/6))^5 = 7^(5/6)

2. **Rewrite the expression:**

(7^(1/3) × 7^(1/2)) ÷ 7^(5/6)

3. **Combine the exponents in the numerator:**

7^((1/3) + (1/2)) ÷ 7^(5/6)

To add 1/3 and 1/2, we need a common denominator, which is 6:

• 1/3 = 2/6
• 1/2 = 3/6

Now, adding them gives us:

2/6 + 3/6 = 5/6

4. **Substituting back into the expression:**

The numerator now becomes:

7^(5/6)

5. **Final step:**

We can now combine the numerator and the denominator:

7^(5/6) ÷ 7^(5/6)

Using the property of exponents where a^m ÷ a^n = a^(m-n), we get:

7^(5/6 - 5/6) = 7^0

And since any number raised to the power of 0 is 1, the final result is:

1