Find the binomial expansion of x2(x−2y)3 and the term independent of x in the expansion of (x − 3/x)4.

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Here are the solutions. 1. Find the binomial expansion of x²(x−2y)³. Step 1: Expand (x−2y)³. (x−2y)³ = x³ + 3x²(−2y) + 3x(−2y)² + (−2y)³; = x³ − 6x²y + 12xy² − 8y³ Step 2: Multiply by x². x²(x³ − 6x²y + 12xy² − 8y³) = x⁵ − 6x⁴y + 12x³y² − 8x²y³ The final answer is x⁵ − 6x⁴y + 12x³y² − 8x²y³. 2. Find the term independent of x in the expansion of (x − 3/x)⁴. Step 1: General term Tk+1. Tk+1 = 4k (x)^(4−k) (−3/x)^k = 4k (−3)^k x^(4−k) x^(−k) = 4k (−3)^k x^(4−2k); Step 2: For term independent of x, set 4−2k=0. 4−2k = 0 ⇒ k=2; Step 3: Substitute k=2. T₂₊₁ = 42 (−3)² x⁴⁻²⁽²⁾ = 6 · 9 · x⁰ = 54 The final answer is 54.

Find the binomial expansion of x2(x-2y)3 and the term independent of x in the expansion of (x - (3)/(x))4.

Step-by-step solution

Expand $(x-2y)^3$ .

Multiply by $x^2$ .

General term $T_{k+1}$ .

For term independent of $x$ , set $4-2k=0$ .

Substitute $k=2$ .

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More Mathematics Questions

Find the binomial expansion of x2(x-2y)3 and the term independent of x in the expansion of (x - (3)/(x))4.

Step-by-step solution

Expand (x−2y)3(x-2y)^3(x−2y)3.

Multiply by x2x^2x2.

General term Tk+1T_{k+1}Tk+1​.

For term independent of xxx, set 4−2k=04-2k=04−2k=0.

Substitute k=2k=2k=2.