Factorise using identity
(x-a)^3 + (x-b)^3 + (x-c)^3 where x = a+b+c/3
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given,x=(a+b+c)/3
(x-a)³+(x-b)³+(x-c)³=3(x-a)(x-b)(x-c)
=x³-(a+b+c)x²+(ab+bc+ac)x-abc
=x³-3x.x²+(ab+bc+ac)x+abc
=-2x³+(ab+bc+ac)x+abc
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(x-a)³+(x-b)³+(x-c)³=3(x-a)(x-b)(x-c)
=x³-(a+b+c)x²+(ab+bc+ac)x-abc
=x³-3x.x²+(ab+bc+ac)x+abc
=-2x³+(ab+bc+ac)x+abc
please mark me brainliest
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