Math, asked by abamruta912, 19 days ago

Expand : ( a+b+c)³. ​

Answers

Answered by devipriyavasudevan16
0

Answer:

(a+b+c)³=(a³+3a²b+3ab²+b³)+3(a²+2ab+b²)c+3(a+b)c2+c³

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

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

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

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

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

Answered by Adrito2
0

Answer:

a³ + b³ + c³ + 3a²b + 3ab² + 3b²c + 3bc² + 3a²c + 3ac² + 6abc

Step-by-step explanation:

( a + b + c )³

= ( a + b + c )( a + b + c )( a + b + c )

= ( a² + ab + ac + ab + b² + bc + ac + bc + c² )( a + b + c )

                                            [Multiplied two first two terms]

= ( a² + b² + c² + 2ab + 2bc + 2ac )( a + b + c )

= a³ + ab² + ac² + 2a²b + 2abc + 2a²c + a²b + b³ + bc² + 2ab² + 2b²c + 2abc + a²c + b²c + c³ + 2abc + 2bc² + 2ac²

                                                    [Multiplied rest two terms]

= a³ + b³ + c³ + 3a²b + 3ab² + 3b²c + 3bc² + 3a²c + 3ac² + 6abc

Ans:- a³ + b³ + c³ + 3a²b + 3ab² + 3b²c + 3bc² + 3a²c + 3ac² + 6abc

Hope it helps you =)

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