Solution
$$\ x^a.x^b.x^c=1$$
$$\ x^{a+b+c}=x^0$$
$$=$$> $$a+b+c=0$$
$$a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$$
$$a^3+b^3+c^3-3abc=0$$
$$a^3+b^3+c^3=3abc$$
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