Given A=⎡⎢⎣abcbcacab⎤⎥⎦
Here, A=AT
Also, |A|=a(bc−a2)−b(b2−ac)+c(ab−c2)
⇒|A|=3abc−a3−b3−c3
⇒|A|=3−a3−b3−c3 (∵abc=1 given)
Also, given AAT=I
⇒A2=I
⇒|A2|=1
⇒|A|=±1
⇒3−a3−b3−c3=−1 (∵ a, b,c are positive real numbers and 3abc−a3−b3−c3=−(a+b+c)(a2+b2+c2−ab−bc−ac))
⇒a3+b3+c3=4