If a = 73, b = 74 and c = 75, then what is the value of $$a^{3} + b^{3} + c^{3}$$ - 3abc?

Given : a = 73, b = 74 and c = 75

We know that, $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

Multiplying both sides by 2, we get :

=> $$2(a^3+b^3+c^3-3abc)=(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)$$

=> $$2(a^3+b^3+c^3-3abc)=(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$$

Substituting values of a,b and c

=> $$a^3+b^3+c^3-3abc=\frac{1}{2}(73+74+75)[(73-74)^2+(74-75)^2+(75-73)^2]$$

= $$\frac{1}{2}(222)(1+1+4)$$

= $$222\times3=666$$

=> Ans - (C)