Let a, b, c ∈ R be all non-zero and satisfies a ^ 3 + b ^ 3 + c ^ 3...

∵ {A}^{T}A = I$</p><p>$⇒ \left[$\begin{cases}a & b & c \\ b & c & a \\ c & a & b\end{cases}$\right]\left[$\begin{cases}a & b & c \\ b & c & a \\ c & a & b\end{cases}$\right] = \left[$\begin{cases}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>$⇒ \left[$\begin{cases}{a}^{2} + {b}^{2} + {c}^{2} & ab + bc + ac & ab + bc + ac \\ ab + bc + ac & {a}^{2} + {b}^{2} + {c}^{2} & ab + bc + ac \\ ab + bc + ac & ab + bc + ac & {a}^{2} + {b}^{2} + {c}^{2}\end{cases}$\right] = \left[$\begin{cases}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>On comparing each element both sides, we get</p><p>${a}^{2} + {b}^{2} + {c}^{2} = 1 \text{ and } ab + bc + ca = 0 . . . . . . . . . \left(i\right)$</p><p>We know that${a}^{3} + {b}^{3} + {c}^{3} - 3abc = \left(a + b + c\right)\left({a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ac\right)$.</p><p>$⇒ 2 - 3abc = \left(a + b + c\right)\left(1 - 0\right)$(from equation \left(i\right)$)

$⇒ 2 - 3abc = \left(a + b + c\right) . . . . . . . . . \left(ii\right)$

Now, we also know that ${\left(a + b + c\right)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2\left(ab + bc + ca\right)$.

$⇒ {\left(a + b + c\right)}^{2} = 1 + 2\left(0\right)$ (from equation \left(i\right)$)</p><p>$⇒ a + b + c = \pm 1$</p><p>Putting it in equation \left(ii\right)$, we get

$2 - 3abc = \pm 1$

$⇒ 3abc = 2 ∓ 1$

$⇒ abc = \dfrac{1}{3}\text{or}abc = 1$