1. **State the problem:** We want to evaluate the expression $$\alpha = \left(2 \cdot \left[ 2 \cdot (3^{-1}) - 2^{-1} \right] - \left[ 2 \cdot (-2)^{-1} + 2^{-1} \right]^{-1}\right)^3$$ in the finite field $\mathbb{Z}_5$, which means all calculations are done modulo 5. 2. **Recall important rules:** In $\mathbb{Z}_5$, the elements are $\{0,1,2,3,4\}$ and arithmetic is modulo 5. - The inverse $a^{-1}$ of an element $a$ satisfies $a \cdot a^{-1} \equiv 1 \pmod{5}$. - Negative numbers are reduced modulo 5, e.g., $-2 \equiv 3 \pmod{5}$. 3. **Find inverses:** - $3^{-1}$ in $\mathbb{Z}_5$: Find $x$ such that $3x \equiv 1 \pmod{5}$. Testing: - $3 \times 2 = 6 \equiv 1 \pmod{5}$, so $3^{-1} = 2$. - $2^{-1}$ in $\mathbb{Z}_5$: Find $x$ such that $2x \equiv 1 \pmod{5}$. Testing: - $2 \times 3 = 6 \equiv 1 \pmod{5}$, so $2^{-1} = 3$. - $(-2)^{-1}$: Since $-2 \equiv 3 \pmod{5}$, this is $3^{-1} = 2$ as above. 4. **Evaluate inner brackets:** - First bracket: $$2 \cdot (3^{-1}) - 2^{-1} = 2 \cdot 2 - 3 = 4 - 3 = 1 \pmod{5}$$ - Second bracket: $$2 \cdot (-2)^{-1} + 2^{-1} = 2 \cdot 2 + 3 = 4 + 3 = 7 \equiv 2 \pmod{5}$$ 5. **Find inverse of second bracket:** - We need $(2)^{-1}$ modulo 5, which is $3$ as found above. 6. **Substitute back:** $$2 \cdot 1 - 3 = 2 - 3 = -1 \equiv 4 \pmod{5}$$ 7. **Raise to the power 3:** $$4^3 = 4 \times 4 \times 4 = 64 \equiv 64 - 60 = 4 \pmod{5}$$ **Final answer:** $$\boxed{4}$$