<s>[INST] Probability of $(a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)=1$ 
A fair die is thrown three times. If $a$, $b$, $c$ are the numbers obtained on the die, then what is the probability that $$(a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)=1$$
  (where $\omega$ is a cube root of unity)

My attempt:
On simplification, $a^{2}+b^{2}+c^{2}=1+(ab+bc+ca)$. I can't figure how to find the number of cases where the condition is satisfied. 
 [/INST] Hint: Since $a,b,c\in\mathbb R$ and $\omega$ is the complex cube root of unity, we can write
$$a^3+b^3+c^3-3abc=(a+b\omega+c\omega^2)(a+c\omega+b\omega^2)(a+b+c)$$
$$(a+b\omega+c\omega^2)(a+c\omega+b\omega^2)=\frac{a^3+b^3+c^3-3abc}{a+b+c}$$
Now, we only have to find out how many integer solutions are there for $a^3+b^3+c^3-3abc=a+b+c\forall 1\leq a,b,c\leq 6$.
On further simplification, 
$$(a^2+b^2+c^2-ab-bc-ca)(a+b+c)=(a+b+c)$$
$$a^2+b^2+c^2-ab-bc-ca=1$$
The total positive integer solutions less than $6$ to the above equation by the total possible dice rolls ($6^3$) will be your required probability.
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