probability that $(a+b\omega+c\omega^2)\cdot (a+b\omega^2+c\omega) = 1$ 
If $\omega$ is a complex cube root of unity with $0<\arg(\omega)<2\pi. $ A fair die is thrown three times
If $a,b,c$ are the numbers obtained on the dice , Then probability that
$(a+b\omega+c\omega^2)\cdot (a+b\omega^2+c\omega) = 1$

$\bf{Attempt:}$ Given $a,b,c\in  \{1,2,3,4,5,6\},$ Then Total number of ways $n(S) = 6^3$
Now for favourable cases $n(A)$
$(a+b\omega+c\omega^2)\cdot (a+b\omega^2+c\omega) = =a^2+b^2+c^2-ab-bc-ca = 1$
$(a-b)^2+(b-c)^2+(c-a)^2 = 2$
Could some help me how to solve it , Thanks

You are looking for the probability of 
$$(a-b)^2+(b-c)^2+(c-a)^2 = 2, \qquad \text{ when }a,b,c \in \{1,2,3,4,5,6\}$$
For the sum of the three squares to be $2$, the only way it can happen is when two of them are $1$ and the third is a $0$. For example, $a=b$, then $c=a \pm 1$. So you are looking for triples of the form $(a,a,a+1)$ or $(a,a,a-1)$ and their cyclic permutations.
For each $a \in \{2,3,4,5\}$, the number of such triplets (including their cyclic permutations) is $24$, then for $a=1$ we can only have $(a,a,a+1)$ and its permutations, likewise for $a=6$ we can only have $(a,a,a-1)$ and its permutations. Therefore  
Thus the probability is $\frac{24+3+3}{6^3}=\frac{30}{6^3}$. 
