What is the quotient of the 6th root of 7 over the cube root of 7

Answer 1

Irrational roots translate to fractional exponents. The "sixth root of seven" is the same as 7 raised to the power of 1/6. The "cube root of seven" is seven raised to the power of 1/3. When exponential values are divided, their exponents Subtract.

So: the quotient of the sixth root of 7 and the cube root of 7 is 7^(1/6) / 7^(1/3)

= 7 ^(1/6 - 1/3)

= 7 ^ (-1/6)

= 1/(7^(1/6)) (remember that a negative exponent becomes positive in the denominator)

7^(1/6) = 1.383 and 1/1.383 = 0.72303

Answer 2

7^(1/6)/7^(1/3)

7^[(1/6)-(1/3)]

7^[1/6)-(2/6)]

7^(-1/6)

1/7^(1/6)

multiply both terms by 7^(5/6) to rationalize the denominator

7^(5/6)/(7^[1/6])(7^[5/6])

7^(5/6)/7