Answer
$(4m+(-n))^4 = 256m^4-256m^3n+96m^2*n^2-16mn^3+n^4$
Work Step by Step
$(a+b)^n = a^n + n/1!* a^{n-1}*b^1 + n(n-1)/2!* a^{n-2}*b^2 + n(n-1)(n-2)/3!* a^{n-3}*b^3 + . . . + b^n$
$a=4m$
$b=-n$
$n=4$
$(4m+(-n))^4 = (4m)^4 + 4/1!* (4m)^{4-1}*(-n)^1 + 4(4-1)/2!* (4m)^{4-2}*(-n)^2 + 4(4-1)(4-2)/3!*(4m)^{4-3}*(-n)^3 + (-n)^4$
$(4m+(-n))^4 = 256m^4 + 4*(4m)^{3}*-n + 4(3)/2*(4m)^{2}*n^2 + 4(3)(2)/6*(4m)^{1}*-n^3 +n^4$
$(4m+(-n))^4 = 256m^4 + 4*64m^3*-n + 6*(16m^2)*n^2 + 4*4m*-n^3+n^4$
$(4m+(-n))^4 = 256m^4-256m^3n+96m^2*n^2-16mn^3+n^4$
