Supposing that you need to expand the binomial, you should use Binomial Theorem such that:

`(a+b)^3 = C_3^0*a^3*b^0 + C_3^1*a^2*b^1 + C_3^2*a*b^2 + C_3^3*a^0*b^3`

Notice that the exponents of a descends from 3 to 0 and the exponents of b ascends from 0 to 3.

The coefficients of each term of expansion are `C_3^0,C_3^1,C_3^2,C_3^3`

You sould remember that `C_n^0 = C_n^n = 1 ` and `C_n^1 = C_n^(n-1) = n` such that:

`C_3^0 = C_3^3 = 1`

` C_3^1 = C_3^2 = 3`

Substituting 1 for `C_3^0,C_3^3` and 3 for `C_3^1, C_3^` 2 yields:

`(a+b)^3 = a^3 + 3a^2*b + 3ab^2 + b^3`

You should factor out 3ab in the group `3a^2*b + 3ab^2` such that:

`(a+b)^3 = a^3 + b^3 + 3ab(a+b)`

**Hence, expanding the cube of binomial yields `(a+b)^3 = a^3 + b^3 + 3ab(a+b).` **

This Solution is for 6-7 th grade students :

(a+b)^3 = (a+b)*(a+b)*(a+b)

= (a+b)^2 *(a+b) [ Assuming one knows the expansion of (a=b)^2 ]

= (a^2 + 2*a*b +b^2) *(a+b) [Using (a+b)^2=( a+b)*(a+b)=a^2 +ab + ab + b^2) = a^2+2ab+b^2]

= a^3+2.a^2.b+a.b^2+a^2.b+2.a.b^2+b^3

= a^3 +2.a^2. b+a^2. b + 2.a.b^2+a.b^2 + b^3

= a^3 + 3.a^2.b + 3.a.B^2 + b^3

Hence, **(a + b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3 Answer**