Suppose, a, b, c are real numbers such that abc = 1. If the matrix   $A=\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]$ is such that ${\mathrm{AA}}^{\mathrm{T}}=\mathrm{I}$  then
the value of a³ + b³ + c³ is

We can see that A′ = A.
Therefore, A′A = I
⇒ A² = I          ⇒      |A²| = |I|
⇒ |A²| = 1        ⇒      |A| = ±1
Therefore, a³ + b³ + c³ – 3abc = ± 1
⇒ a³ + b³ + c³ = 2, 4.
The correct option is (B) and (D)