Answer
$(2x-3)^4 = 16x^4 -96x^3+216x^2-216x+81$
Work Step by Step
$(2x-3)^4$
$(a+b)^n = a^n + n/1!* a^{n-1}*b^1 + n(n-1)/2!* a^{n-2}*b^2 + n(n-1)(n-2)/3!* a^{n-3}*b^3 + . . . + b^n$
$a=2x$
$b=-3$
$n=4$
$(2x+ (-3))^4 = (2x)^4 + 4/1!* (2x)^{4-1}*(-3)^1 + 4(4-1)/2!* (2x)^{4-2}*(-3)^2 + 4(4-1)(4-2)/3!* (2x)^{4-3}*(-3)^3 + 4(4-1)(4-2)(4-3)/4! *(2x)^{4-4}*(-3)^4$
$(2x+(-3))^4 = 16x^4 + 4* (2x)^3*-3+ 4(3)/2*(2x)^{ 2}*9+ 4(3)(2)/6*(2x)^1*-27 + 4(3)(2)(1)/4*3*2*1*(2x)^{0}*81$
$(2x+(-3))^4 = 16x^4 + 4*8x^3*-3+ 6*4x^2*9+4*2x*-27 +81$
$(2x+(-3))^4 = 16x^4 -96x^3+24x^2*9+4*-54x+81$
$(2x+(-3))^4 = 16x^4 -96x^3+216x^2-216x+81$
