Find the value of a3 + b3 + c3 – 3abc when a = 225, b = 236, c = 227.
Answers
Answered by
4
we know that a^3+b^3+c^3-3abc= (a+b+c) 1/2{ (a-b)^2+(b-c)^2+(c-a)^2}
On substituting the values in the formula,
(225 + 236 + 227) * 1/2 (225-236)^2 + (236-227)^2 + (227-225)^2)
= 2034
On substituting the values in the formula,
(225 + 236 + 227) * 1/2 (225-236)^2 + (236-227)^2 + (227-225)^2)
= 2034
Similar questions