can you please simplify these math problems with radicals out for me?

Question

show all steps please because i don't understand this

1. square root of 64 times the 4th root of 64

2. 6th root of 5 times the 3rd root of 5

3. 6th root of 8 times the 6th root of 2


thanks

Answer 1

SQRT(64) * 64^(1/4)
= 64^(1/2) * 64^(1/4) [add exponents 1/2 + 1/4]
= 64^(3/4)
The fourth root of 64 cubed.
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5^(1/6) * 5^(1/3) [add exponents 1/6+1/3 = 1/6+2/6 = 3/6 = 1/2]
5^(1/2)
The square root of 5.
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8^(1/6) * 2^(1/6) [same exponent, mult. bases]
16^(1/6)
The sixth root of 16.
-----

Answer 2

1. 64^(1/2) * 64^(1/4) = 64^(3/4)

2. 5^(1/6) * 5^(1/3) = 5^(3/6) = 5^(1/2)

3. 8^(1/6) * 2^(1/6), 8 = 2*2*2 = 2^3 so:

2^(3*1/6) = 2^(1/2)

Answer 3

1)

[(64)^1/2][(64)^1/4]

= 1/2 + 1/4 = 3/4 therefore

= (64)^3/4

2)

[(5)^1/6][(5)^1/3]

= 1/6 + 1/3 = 3/6 or 1/2 therefore

= (5)^1/2

3)

[(8)^1/6][(2)^1/6]

= [(8)(2)]^1/6

= (16)^1/6

Answer 4

a million) get rid of the unconventional interior the denominator by making use of multiplying the fraction by making use of a million. the particular "style" of a million to apply is the sq. root of three divided by making use of the sq. root of three. 2) The denominator of the fraction would be sq. root of three times the sq. root of three, that's 3. 3) The numerator stands out as the sq. root of three times the sq. root of five, that's the sq. root of 15.

Answer 5

1. sqrt(64*4root(64))
=sqrt(32*4root(64*4))
=sqrt(32*4)
=sqrt(128)
=sqrt(2^7)
=8sqrt(2)