Factorise (a+b-2c)3+(b+c-2a)3+(c+a-2b)3
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3
=(a+b-2c)3+(b+c-2a)3+(c+a-2b)3
=3a +3b-6c+3b+3c-6a+3c+3s-6b
=0
=3a +3b-6c+3b+3c-6a+3c+3s-6b
=0
Answered by
2
3 (a + b - 2c) + 3 (b + c - 2a)+ 3 (c + a - 2b) = 12a + 12b + 12c = 12 (a + b + c)
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