First write out Pascal's triangle as far as the row that reads #1#, #3#, #3#, #1#. This gives you the coefficients of the binomial expansion:
#(a+b)^3 = a^3+3a^2b+3ab^2+b^3#
In the case of #(x+2)# and #(x+3)#, we have the added complication of powers of #2# or powers of #3#, which need to be multiplied by #1#, #3#, #3#, #1# to get the coefficients of #(x+2)^3# and #(x+3)^3#
#color(white)()#
#(x+2)^3 = x^3+(3*x^2*2)+(3*x*2^2)+2^3#
#= x^3+6x^2+12x+8#
#color(white)()#
#(x+3)^3 = x^3+(3*x^2*3)+(3*x*3^2)+3^3#
#= x^3+9x^2+27x+27#
#color(white)()#
To multiply #(x^3+6x^2+12x+8)(x^3+9x^2+27x+27)#, I will write out a #4xx4# grid of products of pairs of coefficients - one from each cubic - and sum the diagonals:
#color(white)(000000)"|"color(white)(00)1color(white)(0000)6color(white)(000)12color(white)(0000)8#
#"---------|-----------------------------"#
#color(white)(000)1color(white)(00)"|"color(white)(00)color(red)(1)color(white)(0000)color(green)(6)color(white)(000)color(blue)(12)color(white)(0000)color(purple)(8)#
#color(white)(000000)"|"#
#color(white)(000)9color(white)(00)"|"color(white)(00)color(green)(9)color(white)(000)color(blue)(54)color(white)(00)color(purple)(108)color(white)(000)color(brown)(72)#
#color(white)(000000)"|"#
#color(white)(00)27color(white)(00)"|"color(white)(0)color(blue)(27)color(white)(00)color(purple)(162)color(white)(00)color(brown)(324)color(white)(00)color(teal)(216)#
#color(white)(000000)"|"#
#color(white)(00)27color(white)(00)"|"color(white)(0)color(purple)(27)color(white)(00)color(brown)(162)color(white)(00)color(teal)(324)color(white)(00)color(orange)(216)#
#color(white)()#
Hence:
#((x+2)(x+3))^3#
#=(x+2)^3(x+3)^3#
#= (x^3+6x^2+12x+8)(x^3+9x^2+27x+27)#
#=color(red)(1)x^6+color(green)((6+9))x^5+color(blue)((12+54+27))x^4+color(purple)((8+108+162+27))x^3+color(brown)((72+324+162))x^2+color(teal)((216+324))x+color(orange)(216)#
#=color(red)(1)x^6+color(green)(15)x^5+color(blue)(93)x^4+color(purple)(305)x^3+color(brown)(558)x^2+color(teal)(540)x+color(orange)(216)#
