Question

Let $a,b,c\in R$ be all non-zero and satisfy $a^3+b^3+c^3=2$. If the matrix $A=\left(\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right)$ satisfies $A^{T}A=I$, then the value of $abc$ can be:

• A. $-\frac{1}{3}$
• B. $3$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is D $\frac{1}{3}$

$a^3+b^3+c^3=2$
$A^{T}A=I$
$\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\Rightarrow a^2+b^2+c^2=1$
$ab+bc+ca=0$

Now ${(a+b+c)}^2=\sum a^2+2\sum ab$

${\left(\sum a\right)}^2=1+0\Rightarrow {\left(\sum a\right)}^2=1\Rightarrow \sum a=\pm 1$

Now $\sum a^3-3abc=(\sum a)\left(\sum a^2-\sum ab\right)$
$2-3abc=\pm 1$