prove that (a+b+c) ³-a³-b³-c³=(a+b)(b+c) (c+a)
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Answered by
17
Answer:
(a+b+c)³-a³-b³-c³
=(a+b)³+3(a+b)²c+3(a+b)c²+c³-a³-b³-c³
=a³+3a²b+3ab²+b³+3(a²+2ab+b²)c+3ac²+3bc²+c³-a³-b³-c³
=3a²b+3ab²+3a²c+6abc+3b²c+3ac²+3bc²
=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)
3(a+b)(b+c)(c+a)
=3(ab+b²+ac+bc)(c+a)
=3(abc+b²c+ac²+bc²+a²b+ab²+a²c+abc)
=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)
LHS=RHS
Answered by
2
Answer:
Question wrong ha
Step-by-step explanation:
3(a+b)(b+c) (c+a) ho ga
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