Question

3 √-32xy^6z^8
need help please

Answer 1

\[\sqrt[3]{-32xy ^{6}z ^{8}}\]

\[\sqrt{4} = 2^(2-2) = 2^1 = 2\]

Answer 2

Treat the cube root s something you subtract the exponents from.

\[\sqrt[3]{9^{5}} = 9^(5-3) = 9^2 = 81\]

Answer 3

the x has an exponent of 1

Answer 4

So it would be:

-32x^(1-3) y^(6-3) z^(8-3)