Here's how we can prove $3=-3$.

Step 1: $\sqrt{3^2}=\sqrt{(-3)^2}$

Step 2: $\sqrt{3 \times 3}=\sqrt{-3 \times -3}$

Step 3: $\sqrt{3} \times \sqrt{3}=\sqrt{-3} \times \sqrt{-3}$

Step 4: $3^{\frac{1}{2}} \times 3^{\frac{1}{2}}= 3^{\frac{1}{2}}i \times 3^{\frac{1}{2}}i$

Step 5: $3^{\frac{1}{2}+\frac{1}{2}}=3^{\frac{1}{2}+\frac{1}{2}} \times i^2$

Hence, $3=-3$.

But we know that $3 \ne -3$. In which step has the error been committed?

**Clarification:** $i=\sqrt{-1}$.

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