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