what is the formula of (a+b+c) whole cube

(a+b+c)³ =  a³ + b³ + c³ + 3(a+b)(b+c)(a+c)

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

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

{a+b+c } ^3 = a^3+b^3+c^3 +3[a+b][b+c][a+c] 

a+b+c whole cube = a cube +b cube +c cube +3(ab) + 3(bc) +3(ac)

( a + b + c )³ =( a + b + c )( a2 + b2 + c2 - ab - bc - ca ) + 3 abc
We transferred the -3 abc to the other side as we required.

wreerhtrj

( a + b + c )³ = a³ + b³ + c³+ 3a²b+3ab² + 3a²c + 3ac² + 3b²c + 3bc² + 6abc

Give me A proper answer

(A+B+C)³=(a+b+c)(A2+b2+c2- ab- bc- ca)

a³+b³+3a²b+3ab²

a^3 +b^3+c^3+3(ab)+3(bc)+3(ca)

(a+b+c)³=(a³+b³+c³)+3a(ab+ac+bc)+ 3b(ab+bc+ac) +3c(ac+bc+ab)−3abc

loll

Let me help you with this formula in detail.
(a + b + c)³ = a³ + b³ + c³ + 3 (a +b) (b + c) (a+ c)
Proof:
(a + b + c)³ = a³ + b³ + c³ + 3 (a +b) (b + c) (a+ c)
It can be written as
(a + b + c)³ - a³ - b³ - c³ =  3 (a +b) (b + c) (a+ c)   ......... (1)
Consider the L.H.S of equation (1),
 (a + b + c)³ - a³ - b³ - c³
=  a³ + b³ + c³ + 3 ab (a + b) + 3 bc (b + c) + 3 ac (a + c) +6 abc - a³ - b³ - c³
= 3 ab (a + b) + 3 bc (b + c) + 3 ac (a + c) +6 abc
= 3 [ ab (a + b) + bc (b + c) + ac (a + c) + 2 abc ]
= 3 [ ab (a + b) + b²c + bc² + abc + a²c + ac² + abc ]
= 3 [ ab (a + b) + (abc + b²c) + (abc + a²c) + (bc² + ac²) ]
= 3 [ ab (a + b) + bc (a + b) + ac (a + b) + c² (a + b) ]
= 3 [ (a + b) (ab + bc + ac + c²) ]
= 3 [ (a + b) { (c² + bc) + ( ab + ac) } ]
= 3 [ (a + b) { c ( b + c ) + a ( b + c ) } ]
= 3 (a + b) ( b + c) ( a + c )
which is equal to R.H.S of equation (1).
Thus proved.
Hope it will help you.
Thanks.

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

(a+b+c)³ =  a³ + b³ + c³ + 3(a+b)(b+c)(a+c)

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(a+b+c)³ =  a³ + b³ + c³ + 3(a+b)(b+c)(a+c)

(a + b + c) ^{3 =} (a + b + c)(a²+ b²+ c² + 2ab + 2ac  + 2bc )
 

(A+B+C)³ = A³+B³+C³+3(A+B)(B+C)(C+A)

a+b+c whole cube = a cube +b cube +c cube +3(ab) + 3(bc) +3(ac)

(a+b+c)³ =  a³ + b³ + c³ + 3(a+b)(b+c)(a+c)

ok
 

(a+b+c)^3  =a^3+ b^3 +c^3  +3ab +3bc +3 cd