With the Pascal's triangle, it's easy to find every binomial expansion :
Each term, of this triangle, is the result of the sum of two terms on the top-line. (example in red)
11
1. 11.1
color(blue)(1. 2. 1)1.2.1
1. color(red)3 . color(red)3. 11.3.3.1
1. 4. color(red)6. 4. 11.4.6.4.1
...
More, each line has the information of one binomial expansion :
The 1st line, for the power 00
The 2nd, for the power 11
The 3rd, for the power 22...
For example : (a+b)^2(a+b)2 we will use the 3rd line in blue following this expansion :
(a+b)^2 = color(blue)1*a^2*b^0 + color(blue)2*a^1*b^1 + color(blue)1*a^0*b^2(a+b)2=1⋅a2⋅b0+2⋅a1⋅b1+1⋅a0⋅b2
Then : (a+b)^2 = a^2 + 2ab + b^2(a+b)2=a2+2ab+b2
To the power 33 :
(a+b)^3 = color(green)1*a^3*b^0 + color(green)3*a^2*b^1 + color(green)3*a^1*b^2 + color(green)1*a^0*b^3(a+b)3=1⋅a3⋅b0+3⋅a2⋅b1+3⋅a1⋅b2+1⋅a0⋅b3
Then (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3(a+b)3=a3+3a2b+3ab2+b3
So here we have color(red)(a=2x)a=2x and color(blue)(b=3)b=3 :
And (2x+3)^3 = color(red)((2x))^3 + 3*color(red)((2x))^2*color(blue)3 + 3*color(red)((2x))*color(blue)3^2 + color(blue)3^3(2x+3)3=(2x)3+3⋅(2x)2⋅3+3⋅(2x)⋅32+33
Therefore : (2x+3)^3 = 8x^3 + 36x^2 + 54x + 27(2x+3)3=8x3+36x2+54x+27
