ALGEBRA QUESTIONS

IF a = 997, b = 998, c = 999,

 THEN,

 FIND, a^3 + b^3 + c^3 - 3abc

 WITH THE FORMULA OF,

a^3 + b^3 + c^3 - 3abc = (a + b + c)/2{(a-b)^2 + (b - c)^2 + (c - a)^2}

 (997 + 998 + 999)/2{(997 - 998)^2 + (998 - 999)^2 + (999 - 997)^2}

 2994/2 * 1 + 1+ 4

 2994/2 * 6

 2994 * 3

 8982

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