Let a, b, c ∈R be such that a+b+c>0 and abc=2. Let A=a    b    cb    c    ac    a    b If A2=I, then value of a3+b3+c3 is

# Let  be such that $a+b+c>0$ and $abc=2.$ Let $A=\left[\begin{array}{l}a b c\\ b c a\\ c a b\end{array}\right]$ If ${A}^{2}=I,$ then value of ${a}^{3}+{b}^{3}+{c}^{3}$ is

1. A

7

2. B

2

3. C

0

4. D

-1

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### Solution:

${A}^{2}=\left[\begin{array}{l}a b c\\ b c a\\ c a b\end{array}\right]\left[\begin{array}{l}a b c\\ b c a\\ c a b\end{array}\right]=\left[\begin{array}{ccc}\alpha & \beta & \beta \\ \beta & \alpha & \beta \\ \beta & \beta & \alpha \end{array}\right]$
where $\alpha ={a}^{2}+{b}^{2}+{c}^{2},$
$\beta =bc+ca+ab.$
As ${A}^{2}=I,$ we get
${a}^{2}+{b}^{2}+{c}^{2}=\alpha =1$
$bc+ca+ab=0$
Now, $\left(a+b+c{\right)}^{2}=\alpha +2\beta =1$

We have
$\begin{array}{l}{a}^{3}+{b}^{3}+{c}^{3}-3abc\\ =\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-bc-ca-ab\right)\\ =1\\ ⇒{a}^{3}+{b}^{3}+{c}^{3}=7\end{array}$  