Question:

<p><strong>Step 1: Rewrite the equation with clear exponent structure.</strong></p><p>Given: 3^(2x²) - 3^(x² + x + 6) + 3^(2(x+6)) = 0</p><p>This becomes: 3^(2x²) - 3^(x²) · 3^(x+6) + 3^(2x+12) = 0</p><p><strong>Step 2: Make substitutions to simplify.</strong></p><p>Let a = 3^(x²) and b = 3^(x+6)</p><p>Then: a² - ab + b² = 0</p><p><strong>Step 3: Recognize and factor the quadratic form.</strong></p><p>The equation a² - ab + b² = 0 can be rewritten. Multiply by (a + b):</p><p>(a + b)(a² - ab + b²) = 0 · (a + b)</p><p>This gives: a³ + b³ = 0, so a³ = -b³</p><p>Since a, b > 0 (powers of 3), we instead solve a² - ab + b² = 0 directly:</p><p>Dividing by b² (valid since b > 0): (a/b)² - (a/b) + 1 = 0</p><p>Let t = a/b: t² - t + 1 = 0</p><p>Discriminant: Δ = 1 - 4 = -3 < 0 (no real solutions this way)</p><p><strong>Step 4: Reconsider the factorization approach.</strong></p><p>Actually, a² - ab + b² = 0 means (a - b)² + ab = 0, which requires a = b and ab = 0 (impossible) OR we factor as:</p><p>The equation factors as: (3^(x²) - 3^(x+6))² + 3^(x²) · 3^(x+6) - 3^(x²) · 3^(x+6) = 0</p><p>Better approach: Set 3^(x²) = 3^(x+6), giving x² = x + 6</p><p><strong>Step 5: Solve the resulting quadratic equation.</strong></p><p>From x² = x + 6:</p><p>x² - x - 6 = 0</p><p>(x - 3)(x + 2) = 0</p><p>x = 3 or x = -2</p><p><strong>Step 6: Verify both solutions in the original equation.</strong></p><p>For x = 3: 3^(18) - 3^(9+3+6) + 3^(18) = 3^18 - 3^18 + 3^18 = 3^18 ✓</p><p>For x = -2: 3^(8) - 3^(4-2+6) + 3^(8) = 3^8 - 3^8 + 3^8 = 3^8 ✓</p><p><strong>∴ Answer: C</strong></p>