Question

What is 2x (to the second power)y (to the fourth power ) times 4x (squared) y (to the fourth power) times 3x over 3x (negative third power) y(squared)

Answer 1

`2x^2y^4*4x^2y^4*(3xy^2)/(3x^(-3))= 8x^4y^8* (3xy^2)/(1)*(3x^3)/(1)= \\\\ =8x^4y^8*9x^4y^2= \\\\ =\boxed{72x^8y^(10)}`