What is the cube of (a+b+c)?

Question

(a+b+c)3(to the power 3)

Answer 1

In TeX:

(a+b+c)^{3} = \sum_{i,j=0}^{3} \frac{3!}{i! j! (3-i-j)!} a^{i} b^{j} c^{3-i-j}

For the expansion of (a + b + c)^{n}, replace 3 with n everywhere in the above formula.

Answer 2

Eulercrosser has it right except that he forgot the abc term. If you substitute a = b = c = 1, you get (1+1+1)^3 which is 3^3 or 27. So the coefficients should add up to 27. Eulercrosser's coefficients add to 21. Each term of the answer consists of picking a letter from {a,b,c} three times (for example, bca or aab), so this becomes a combinatorial problem. The missing term is +6abc:

a^3+3(a^2)b+3(a^2)c+ b^2+3(b^2)a+3(b^2)c +c^2+3(c^2)a+3(c^2)b + 6abc

Answer 3

(a+b+c)^3

Answer 4

(a+b+c)^3

Answer 5

(a+b+c)(a+b+c)(a+b+c)

The product of the first 2 factors is

a^2+2ab+2ac+b^2+2bc+c^2

Multiplying this result by the third factor gives

a^3 + 3a^2(b) + 3a^2(c) + 3ab^2 + 6abc + 3ac^2 + b^3
+ 3b^2(c) + 3bc^2 + c^3

Answer 6

(a+b+c)^3= ((a+b)+c)^3= (a+b)^3 +3*(a+b)^2*c+3*(a+b)*c^2 +c^3=.... just continue the process

Answer 7

(a + b + c)^3

(a + b + c)(a + b + c)(a + b + c)

(a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2)(a + b + c)

(a^2 + 2ab + 2ac + 2bc + b^2 + c^2)(a + b + c)

a^3 + a^2b + a^2c + 2a^2b + 2ab^2 + 2abc + 2a^2c + 2abc + 2ac^2 + 2abc + 2b^2c + 2bc^2 + ab^2 + b^3 + b^2c + ac^2 + bc^2 + c^3

a^3 + 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + 3b^2c + 3bc^2 + b^3 + c^3

Answer 8

(a+b+c)^3=(a+b+c)(a+b+c)(a+b+c)
=(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc)(a+b+c)
=(a^3 + ab^2 + ac^2 +2a^2b + 2a^2c + 2abc + a^2b + b^3 + bc^2 + 2ab^2 + 2abc + 2b^2c + a^2c + b^2c + c^3 + 2abc + 2ac^2 + 2bc^2)

you'll have to check my arithmetic over, I haven't done any math for 3 years!

Answer 9

a^3 + b^3 + c^3 + 3ab(a+b) + 3bc(b+c) + 3ac(a+c) + 6abc

Answer 10

a3+b3+c3