Prove that (a + b)³ + (b + c)³ + (c + a)³ - 3(a + b)(b + c)(c + a) = 2(a³ + b³ + c³ - 3abc)
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1)Open all brackets
a³+b³+3a²b +3ab²+b³+c³+3b²c +3bc²+c ³+a ³+3a²c+3ac²-3(abc+a²b+ac ²+a ²c+b ²c+b ²a+bc²+abc)
=2a³+2b³+2c³+3a²b+3 ab²+3 a²c+3 ac²+3bc²+3 b²c - 3a²b+3 ab²+3 a²c+3 ac²+3 b²c+3 bc²
=2a³+2b³+2c³-6abc
=2(a³+b³+c³+3abc)
a³+b³+3a²b +3ab²+b³+c³+3b²c +3bc²+c ³+a ³+3a²c+3ac²-3(abc+a²b+ac ²+a ²c+b ²c+b ²a+bc²+abc)
=2a³+2b³+2c³+3a²b+3 ab²+3 a²c+3 ac²+3bc²+3 b²c - 3a²b+3 ab²+3 a²c+3 ac²+3 b²c+3 bc²
=2a³+2b³+2c³-6abc
=2(a³+b³+c³+3abc)
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12
expand old the brackets and take 2 in common
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