Find the value of 53^3 using the identity (x+y)^3=x^3+3x^2y+3xy^2+y^2

(53)^3  =  (50 + 3)^3

Using the binomial expansion  of (x + y) ^3, we can write

(x + y)^3   =  1x^3  + 3x^2y + 3xy^2  + 1y^3        where the coefficients come from the 3rd row of Pascal's triangle  [ the first row = 1  =  row "0"  ]   ......so....substituting for x and y, we have

(50 + 3)^3  =  1*50^3 + 3(50)^2(3) + 3(50)(3)^2  + 1*3^3   =

125,000 +  22500 + 1350 + 27  =

148,877

WHAT IS IT THAT YOU DO NOT GET??????

WAS ANSWERED YESTERDAY; HERE IT IS AGAIN. EXPLAIN WHAT YOU DON'T UNDERSTAND.

(x + y)3 = x3 + 3x2y + 3xy2 + y3.
(50 + 3)^3=x^3 + 3*x^2*y +3*x*y^2 + y^3

148,877 =50^3 +9*2,500  + 1,350 + 27

148,877=125,000 +22,500 + 1,350 + 27