Question

prove that . please guys

Answer 1

Answer:

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

Step-by-step explanation:

Given Equation is (a + b + c)³ - a³ - b³ - c³

= a³ + b³ + c³ + 3ab(a + b) + 3bc(b + c) + 3ac(a + c) + 6abc - a³ - b³ - c³

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

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

= 3[ab(a + b) + b²c + bc² + a²c + ac² + abc + abc]

= 3[ab(a + b) + b²c + abc + bc² + a²c + ac² + abc]

= 3[ab(a + b) + bc(a + b) + c²(a + b) + ac(a + b)]

= 3(a + b)[ab + bc + c² + ac]

= 3(a + b)[bc + c² + ab + ac]

= 3(a + b)[c(b + c) + a(b + c)]

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

Hope it helps!

Answer 2

( a + b + c )³

Use : ( x + y )³ = x³ + y³ + 3 x y ( x + y )

Take x as a + b and y as c

==> [ ( a + b )³ + c³ + 3 c ( a + b ) ( a + b + c) ]

Now expanding : ( a + b )³

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

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

Now : ( a + b + c )³ - a³ - b³ - c³

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

==> ( a + b )( 3 ab + 3 c( a + b + c ) ) [  take ( a + b ) common ]

==> ( a + b )( 3 ab + 3 ac + 3 bc + 3 c² )

==> 3 ( a + b )( ab + ac + bc + c² )

==> 3 ( a + b )( a ( b + c ) + c ( b + c ) ) [ taking a and c common ]

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

[ Proved ]

#JISHNU#

Hope it helps mate :-)

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