factorise (2a-b-c)^3 + (2b-c-a)^3 + (2c-a-b) ^3
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We know that, if x+y+z=0; then x 3+y 3+z 3=3 x y z
We know that, if x+y+z=0; then x 3+y 3+z 3=3 x y z(2a-b-c)+(2b-c-a)+(2c-a-b)=0
We know that, if x+y+z=0; then x 3+y 3+z 3=3 x y z(2a-b-c)+(2b-c-a)+(2c-a-b)=0∴ (2a-b-c)3+(2b-c-a)3+(2c-a-b)3= 3(2a-b-c)(2b-c-a)(2c-a-b)
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