If a = 101, b = 99 and c = 100, what is a³ + b³ + c³ − 3abc?

900. a+b+c = 300 and a^2+b^2+c^2-ab-bc-ca = 1/2 [(a-b)^2+(b-c)^2+(c-a)^2 ] = 1/2(4+1+1) = 3. Value = 300 × 3 = 900.

If $a = 101, b = 99$ and $c = 100$ , what is $a^{3} + b^{3} + c^{3} - 3 ab c$ ?

Competition practiceAlgebra

a + b + c = 300

and

a^{2} + b^{2} + c^{2} - ab - b c - c a = \frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = \frac{1}{2} (4 + 1 + 1) = 3

. Value

= 300 \times 3 = 900

Answer: D) 900

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