Prove that (x-a)^3 + (x-b)^3 + (x-c)^3 -3 (x-a)(x-b)(x-c) = 0 . When a+b+c =0
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the condition should be a+b+c=3x not a+b+c =0
x-a+x-b+x-c=3x-(a+b+c)=3x-3x=0
hence (x-a)^3 + (x-b)^3 + (x-c)^3 =3 (x-a)(x-b)(x-c)
⇒ (x-a)^3 + (x-b)^3 + (x-c)^3 -3 (x-a)(x-b)(x-c) =0
x-a+x-b+x-c=3x-(a+b+c)=3x-3x=0
hence (x-a)^3 + (x-b)^3 + (x-c)^3 =3 (x-a)(x-b)(x-c)
⇒ (x-a)^3 + (x-b)^3 + (x-c)^3 -3 (x-a)(x-b)(x-c) =0
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