⚡ AZUCATION SHORTCUT:
Substitute $3^{x^2+2x-3} = t$. Equation becomes $t^2 - 12t + 27 = 0$.
Roots are $t=9, 3$.
Powers: $x^2+2x-3 = 2$ and $x^2+2x-3 = 1$.
Product of roots from Eq 1 is $-5$; from Eq 2 is $-4$.
Total Product $= (-5) \times (-4) = \mathbf{20}$.

Step 1: Use Substitution
Let $k = x^2 + 2x - 3$. Rewrite the terms:
$9^k - 4(3^{k+1}) + 27 = 0$
$(3^k)^2 - 12(3^k) + 27 = 0$.

Step 2: Solve the Quadratic in $3^k$
Let $3^k = t$. Then $t^2 - 12t + 27 = 0 \implies (t-9)(t-3) = 0$.
So, $3^k = 9 \implies k = 2$ OR $3^k = 3 \implies k = 1$.

Step 3: Solve for $x$ and find the product
1. $x^2 + 2x - 3 = 2 \implies x^2 + 2x - 5 = 0$. Product of roots ($x_1 x_2$) $= -5$.
2. $x^2 + 2x - 3 = 1 \implies x^2 + 2x - 4 = 0$. Product of roots ($x_3 x_4$) $= -4$.
Product of all values $= (-5) \times (-4) = \mathbf{20}$.

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