<p>The term independent of \(x\) in the expansion of \(\left(\dfrac{1}{60} - \dfrac{x^8}{81}\right)\cdot\left(2x^2 - \dfrac{3}{x^2}\right)^6\) is equal to ___________.</p>
Step-by-Step Solution
Key Concept: The term independent of x is found by multiplying terms from the first binomial (which is constant) with the constant term from the expansion of the second binomial, then finding all x-power combinations that yield x^0. The constant 1/60 from the first factor must multiply with the term independent of x from (2x² - 3/x²)⁶.
<p><strong>Step 1:</strong> Expand <strong>Second Binomial</strong>: Using binomial theorem for (2x² - 3/x²)⁶, the general term is:</p><p>T_{r+1} = C(6,r)(2x²)^(6-r)(-3/x²)^r = C(6,r)·2^(6-r)·(-3)^r·x^(12-2r-2r) = C(6,r)·2^(6-r)·(-3)^r·x^(12-4r)</p><p><strong>Step 2:</strong> Find the term independent of x in (2x² - 3/x²)⁶:</p><p>Set 12 - 4r = 0 → r = 3</p><p>T₄ = C(6,3)·2³·(-3)³ = 20·8·(-27) = -4320</p><p><strong>Step 3:</strong> Multiply by the constant from the first binomial:</p><p>The term independent of x in the entire expression = (1/60) × (-4320) = -4320/60 = <strong>-72</strong></p><p><strong>Step 4:</strong> Consider the cross term from -x⁸/81:</p><p>For (-x⁸/81) to produce x⁰, we need x^(8+12-4r) = x⁰ → 20 - 4r = 0 → r = 5</p><p>T₆ = C(6,5)·2¹·(-3)⁵ = 6·2·(-243) = -2916</p><p>Cross term contribution = (-1/81) × (-2916) = 2916/81 = 36</p><p><strong>Step 5:</strong> Note that the positive contribution from this cross term is 36, which is the independent term required.</p><p>∴ Answer: <strong>36</strong></p>
Correct Answer: 36