<p>Given \(\left(\dfrac{1}{60} - \dfrac{x^8}{81}\right)\left(2x^2 - \dfrac{3}{x^2}\right)^6\), find the coefficient of \(x^0\).</p>
Step-by-Step Solution
Key Concept: The coefficient of x⁰ comes from terms in the binomial expansion of (2x² - 3/x²)⁶ that, when multiplied by 1/60 or -x⁸/81, produce constant terms. You must identify which general term Tₖ₊₁ gives x⁰ after accounting for the factor multiplying the entire expansion.
<p><strong>Step 1:</strong> Find the general term in (2x² - 3/x²)⁶</p><p>T_{k+1} = C(6,k)(2x²)^{6-k}(-3/x²)^k = C(6,k)·2^{6-k}·(-3)^k·x^{12-2k-2k} = C(6,k)·2^{6-k}·(-3)^k·x^{12-4k}</p><p><strong>Step 2:</strong> From the 1/60 part: need x⁰ from binomial, so 12 - 4k = 0 → k = 3</p><p>T₄ = C(6,3)·2³·(-3)³·x⁰ = 20·8·(-27) = -4320</p><p>Coefficient from 1/60 part: (1/60)·(-4320) = -72</p><p><strong>Step 3:</strong> From the -x⁸/81 part: need x⁸ from binomial (so -x⁸/81·x⁸ = -x¹⁶/81... wait, reconsider)</p><p>Actually need: 12 - 4k = 8 → k = 1</p><p>T₂ = C(6,1)·2⁵·(-3)¹·x⁸ = 6·32·(-3)·x⁸ = -576x⁸</p><p>Coefficient from -x⁸/81 part: (-1/81)·(-576) = 576/81 = 64/9... Recalculate: need 12-4k = -8 (impossible) OR recognize -x⁸/81 · T₂ needs adjustment.</p><p>For x⁰ from second part: 8 + (12-4k) = 0 → k = 5</p><p>T₆ = C(6,5)·2¹·(-3)⁵·x⁻⁸ = 6·2·(-243)·x⁻⁸ = -2916x⁻⁸</p><p>Coefficient from -x⁸/81 part: (-1/81)·(-2916) = 36</p><p><strong>Step 4:</strong> Total coefficient of x⁰ = -72 + 36 = -36</p><p>∴ Answer: <strong>-36</strong></p>
Correct Answer: -36