help :)

Question

1.) cube root of 3x^2 multiplied to square root of 2x?

2.) square root of 3+2sqrt2 multiplied to sqrt of 3 - 2sqrt2

3.) sixth root of 12 divided by (sqrt 3 multiplied to cube root of 2)

4.) 4/cuberoot of 16

Answer 1

∛(3x^2) * √(2x)
= 6throot ((3x^2)^2 * (2x)^3)
= 6throot(72x^7)

√(3+2√2) * √(3-2√2)
= √((3+2√2)(3-2√2))
= √(9-8)
= 1

6throot(12) / (√3*∛2)
= 6throot(12) / 6throot(3^3 * 2^2)
= 6throot(12/36)
= 1/6throot(3)

4/∛16
(2^2)/(2^(4/3))
= 2^(2 - 4/3)
= 2^(2/3)
= ∛4
Or, you could do
∛64/∛16
= ∛(64/16)
= ∛4