Find the first three terms in ascending powers of y in the expansion of (1-y)^(1/3) and the coefficient of…

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Find the first three terms in ascending powers of y in the expansion of (1-y)^(1/3) and the coefficient of x⁻³ in the expansion of ((x+1)(2x − 1/x)3).

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Here are the solutions. 3. Find the first three terms in ascending powers of y in the expansion of (1−y)^(1/3). Use the binomial expansion (1+x)ⁿ = 1 + nx + (n(n−1))/(2!)x² +; Here x=−y and n=1/3. Step 1: First term. 1 Step 2: Second term. n(−y) = 1/3(−y) = −1/3y Step 3: Third term. (n(n−1))/(2!)(−y)² = (1/3(1/3−1))/(2) y² = (1/3(−2/3))/(2) y² = (−2/9)/(2) y² = −1/9y² The first three terms are 1 − 1/3y − 1/9y². 4. Find the coefficient of x⁻³ in the expansion of ((x+1)(2x − 1/x)³). Step 1: Expand (2x − 1/x)³. (2x − 1/x)³ = (2x)³ + 3(2x)²(−1/x) + 3(2x)(−1/x)² + (−1/x)³; = 8x³ − 12x + 6/x − (1)/(x³); = 8x³ − 12x + 6x⁻¹ − x⁻³ Step 2: Multiply by (x+1). (x+1)(8x³ − 12x + 6x⁻¹ − x⁻³) = x(8x³ − 12x + 6x⁻¹ − x⁻³) + 1(8x³ − 12x + 6x⁻¹ − x⁻³); = (8x⁴ − 12x² + 6x⁰ − x⁻²) + (8x³ − 12x + 6x⁻¹ − x⁻³) Step 3: Identify the coefficient of x⁻³. The only term with x⁻³ is −x⁻³. The coefficient of x⁻³ is −1. The final answer is −1.

Find the first three terms in ascending powers of y in the expansion of (1-y)^(1/3) and the coefficient of x⁻³ in the expansion of ((x+1)(2x - 1/x)3).

Answer

Step-by-step solution

First term.

Second term.

Third term.

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

Multiply by $(x+1)$ .

Identify the coefficient of $x^{-3}$ .

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