Question

56x7y5, if x+0 and y#.0?

Answer 1

Answer:

A is the answer.

Step-by-step explanation:

We have

$\sqrt[3]{56 {x}^{7} {y}^{5} }$

Factor the radical by taking perfect squares of the real numbers and the exponents.

56 can be made up of perfect 8, x^7 can be made up of x^6 and y^5 can be made up of y^3. So we get

$\sqrt[3]{8 {x}^{6} {y}^{3} \times \sqrt{7xy {}^{2} } }$

Take the cube root of 8x^6y^3.

Which is 2x^2y.

$2 {x}^{2} y \sqrt[3]{7xy {}^{2} }$

A is the answer.

Answer 2

B) answer is there