Question

What's the formula for (a+b+c+d) the whole cube

Answer 1

Answer:a^2+b^2+c^2+d^2+2ac+2ad+2ab+2bc+2bd+2dc

Step-by-step explanation:

Answer 2

Answer:

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.

Step-by-step explanation:

Let us just start with 

(a+b+c)² = a² +b² + c²+2ab+2bc+2ca

=a² +b² + c²+2(ab+bc+ca)

now:

(a+b+c)² (a+b+c)=(a + b + c)³=[a² +b² + c²+2(ab+bc+ca)](a+b+c)

=a²[a+b+c] +b²[a+b+c]+c²[a+b+c] +2[(ab+bc+ca)[a+b+c]]

=a³+a²b+a²c + b²a+b³+b²c + c²a +c²b +c³ + 2ab[a+b+c] +2bc(a+b+c)

+2ca(a+b+c)

=a³+a²b+a²c + b²a+b³+b²c + c²a +c²b +c³ +2a²b+2ab²+2abc +2abc+2b²c +2bc² +2a²c +2abc +2a²c

=a³+b³+c³+6abc +a²b+2a²b+a²c + b²a+b²c+2ab² +c²a+c²b+2c²b

=a³+b³+c³+6abc+3a²[b+c] +3b²(a+c) +3c²(a+b)

=a³+b³+c³+6abc+3[a²[b+c] +b²(a+c) +c²(a+b)]

on further simplifying, we get

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