what is (a+b+c)^3?

Answer 1

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

let b+c = d

=(a+d)^3
so,

(3C0) a^3d^0 + (3C1)a^2d + (3C2) ad^2 + (3C3) a^0d^3

= 1 a^3 + 3a^2d + 3ad^2 + d^3

so, put in the value of d

= 1 a^3 + 3a^2(b+c) + 3a(b+c)^2 + (b+c)^3

= a^3 + 3a^2.b + 3a^2.c + 3a (b^2+c^2+2bc)+ (b^3+c^3+3b^2c+ 3c^b)..

and jsut simply..this last step..

Answer 2

a^3 + 3*(b+c)a^2 + b^3 + 3(a+c)b^2 + 3c^3 + 3(a+b)c^2 +6abc
I used two iterations of Pascal's Triangle.

Answer 3

Set it up like a regular multiplication:

a+b+c
a+b+c
--------- multiply
a^2+ab+ac
ab+b^2+bc
ac+bc+c^2
------------- collect like terms.


Then do it again.

This method works for the product of any two bracketed factors!
Down with foil!

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Answer 5

it is

a^3+b^3+c^3+3a²b+3ab²+3ac²+3a²

Answer 6

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

NOW,do your home-work.

Answer 7

1 + 1 + 1= 3 ANYWAY YOUR FAT

Answer 8

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

Answer 9

(a+b+c)*(a+b+c)^2
(a+b+c)(a²+b²+c²+2ab+2ac+2bc)
a^3+b^3+c^3+3a²b+3ab²+3ac²+3a²c+3b²c+3bc²+6abc