Question

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

Answer 1

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

=(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)]

Answer 2

Answer:

L:H:S:

(a+b+c)³

= {a+(b+c)}³

= a³+(b+c)³+3a(b+c)(a+b+c)

=a³+b³+c³+3bc(b+c)+3a (b+c)(a+b+c)

=a³+b³+c³+3(b+c)(bc+a²+ab+ac)

=a³+b³+c³+3(b+c)(bc+ca+a²+ab)

= a³+b³+c³+3(b+c){c(b+a)+a(a+b)}

=a³+b³+c³+3(a+b)(b+c)(c+a)

= R:H:S

hope it helps ✨✨