Solution:
[tex]\frac{3x^0(2x^3y^2)^4}{(4x^7y^4)^2}[/tex]
[tex]\frac{3x^0(2x^3y^2)^4}{(4x^7y^4)^2} =\frac{3(1)(2x^3y^2)^4}{(4x^7y^4)^2}[/tex] Since, [tex]a^0=1[/tex]
[tex]\frac{3x^0(2x^3y^2)^4}{(4x^7y^4)^2} =\frac{3(2)^4(x^3)^4(y^2)^4}{(4)^2(x^7)^2(y^4)^2}[/tex] Since, [tex](ab)^m=a^mb^m[/tex]
[tex]\frac{3x^0(2x^3y^2)^4}{(4x^7y^4)^2} =\frac{3(16)x^{12}y^{8}}{16x^{14}y^{8}}[/tex] Since, [tex](a^m)^n=a^{mn}[/tex]
[tex]\frac{3x^0(2x^3y^2)^4}{(4x^7y^4)^2} =\frac{3x^{12}y^{8}}{x^{14}y^{8}}[/tex]
[tex]\frac{3x^0(2x^3y^2)^4}{(4x^7y^4)^2} =3x^{12-14}y^{8-8}[/tex] Since, [tex]\frac{a^m}{a^n} =a^{m-n}, a^0=1[/tex]
[tex]\frac{3x^0(2x^3y^2)^4}{(4x^7y^4)^2} =3x^{-2}y^{0}[/tex]
[tex]\frac{3x^0(2x^3y^2)^4}{(4x^7y^4)^2} =\frac{3}{x^2}[/tex]
