Find the coefficient of x⁻³ in the expansion of (x+1)(2x − 1/x)3.

Question

ScanSolve AI Tutor · Accepted Answer

Step 1: Expand the binomial term (2x − 1/x)³. We use the binomial theorem (a+b)ⁿ = n0aⁿ b⁰ + n1aⁿ⁻¹b¹ + n2aⁿ⁻²b² + + nna⁰ bⁿ. Here, a=2x, b=−1/x=−x⁻¹, and n=3. The terms are: Term 1: 30(2x)³(−x⁻¹)⁰ = 1 · (8x³) · 1 = 8x³; Term 2: 31(2x)²(−x⁻¹)¹ = 3 · (4x²) · (−x⁻¹) = −12x²⁻¹ = −12x; Term 3: 32(2x)¹(−x⁻¹)² = 3 · (2x) · (x⁻²) = 6x¹⁻² = 6x⁻¹; Term 4: 33(2x)⁰(−x⁻¹)³ = 1 · 1 · (−x⁻³) = −x⁻³ So, the expansion of (2x − 1/x)³ is 8x³ − 12x + 6x⁻¹ − x⁻³. Step 2: Multiply the expanded form by (x+1) and identify terms with x⁻³. We need to find the coefficient of x⁻³ in the product (x+1)(8x³ − 12x + 6x⁻¹ − x⁻³). This product can be written as: x(8x³ − 12x + 6x⁻¹ − x⁻³) + 1(8x³ − 12x + 6x⁻¹ − x⁻³) Let's examine the terms that will result in x⁻³: From the first part, x(8x³ − 12x + 6x⁻¹ − x⁻³): x · (8x³) = 8x⁴; x · (−12x) = −12x²; x · (6x⁻¹) = 6x¹⁻¹ = 6x⁰ = 6; x · (−x⁻³) = −x¹⁻³ = −x⁻² None of these terms result in x⁻³. From the second part, 1(8x³ − 12x + 6x⁻¹ − x⁻³): 1 · (8x³) = 8x³; 1 · (−12x) = −12x; 1 · (6x⁻¹) = 6x⁻¹; 1 · (−x⁻³) = −x⁻³ This term results in x⁻³, and its coefficient is −1. Step 3: Sum the coefficients of x⁻³. The only term contributing to x⁻³ is −x⁻³ from the second part of the multiplication. Therefore, the coefficient of x⁻³ is −1. The final answer is −1.

Find the coefficient of x⁻³ in the expansion of (x+1)(2x - 1/x)3.

Answer

Step-by-step solution

Expand the binomial term $\left(2x - \frac{1}{x}\right)^3$ .

Multiply the expanded form by $(x+1)$ and identify terms with $x^{-3}$ .

Sum the coefficients of $x^{-3}$ .

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