Question

cube of (ab-bc).....plzzz answer fast​

Answer 1

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.

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Step-by-step explanation:

Answer 2

check the 2 pic

hope this helps you

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