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

A
7
B
2
C
0
D
-1

Solution:

$A^{2} = [\begin{array}{lll} a b c \\ b c a \\ c a b \end{array}] [\begin{array}{lll} a b c \\ b c a \\ c a b \end{array}] = [\begin{array}{ccc} α & β & β \\ β & α & β \\ β & β & α \end{array}]$
where $α = a^{2} + b^{2} + c^{2},$
$β = b c + c a + a b .$
As $A^{2} = I,$ we get
$a^{2} + b^{2} + c^{2} = α = 1$
$b c + c a + a b = 0$
Now, $(a + b + c)^{2} = α + 2 β = 1$
$\Rightarrow a + b + c = 1$
We have
$\begin{array}{l} a^{3} + b^{3} + c^{3} - 3 a b c \\ = (a + b + c) (a^{2} + b^{2} + c^{2} - b c - c a - a b) \\ = 1 \\ \Rightarrow a^{3} + b^{3} + c^{3} = 7 \end{array}$

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

Solution:

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