Correct Answer: $2$

Solution : Given: $\frac{(1.2)^{3} + (0.8)^{3} + (0.7)^{3} – 2.016}{(1.35)[(1.2)^{2} + (0.8)^{2} + (0.7)^{2} – 0.96 – 0.84 – 0.56]}$
We know that the algebraic identity is $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
$=\frac{(1.2)^{3} + (0.8)^{3} + (0.7)^{3} – 3\times1.2\times0.8\times0.7}{(1.35)[(1.2)^{2} + (0.8)^{2} + (0.7)^{2} – 1.2\times0.8– 1.2\times0.7– 0.8\times0.7]}$
$=\frac{(1.2+0.8+0.7)[(1.2)^{2} + (0.8)^{2} + (0.7)^{2} – 1.2\times0.8– 1.2\times0.7– 0.8\times0.7]}{(1.35)[(1.2)^{2} + (0.8)^{2} + (0.7)^{2} – 1.2\times0.8– 1.2\times0.7– 0.8\times0.7]}$
$=\frac{1.2+0.8+0.7}{1.35}=\frac{2.7}{1.35}=2$
Hence, the correct answer is $2$.