If a = 4.36, b = 2.39 and c = 1.97, then the value of a3 - b3 - c3 - 3abc is?

Answer: Option C

$$\begin{array}{rl}& a=4.36\\ & b=2.39\\ & c=1.97\\ & a-b-c\\ & =4.36-2.39-1.97\\ & =0\\ & {\text{a}}^3-b^3-c^3-3abc\\ & =\frac{1}{2}\left(a-b-c\right)\left[{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right]\\ & =0\end{array}$$