The coefficient of x^3 in the expansion of (x | vedclass

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Question

(B) Let $P(x) = (x^2 - x + 1)^{10} (x^2 + 1)^{15}$.We need the coefficient of $x^3$ in the expansion.$(x^2 - x + 1)^{10} = 1 + 10(x^2 - x) + \binom{10

Vedclass · Accepted Answer

$-240$

Vedclass · Answer

(B) Let $P(x) = (x^2 - x + 1)^{10} (x^2 + 1)^{15}$.We need the coefficient of $x^3$ in the expansion.$(x^2 - x + 1)^{10} = 1 + 10(x^2 - x) + \binom{10}{2}(x^2 - x)^2 + \binom{10}{3}(x^2 - x)^3 + \dots$$= 1 + 10x^2 - 10x + 45(x^4 - 2x^3 + x^2) + 120(x^6 - 3x^5 + 3x^4 - x^3) + \dots$$= 1 - 10x + 55x^2 - 90x^3 + \dots$$(x^2 + 1)^{15} = \binom{15}{0} + \binom{15}{1}x^2 + \binom{15}{2}x^4 + \dots = 1 + 15x^2 + \dots$Multiplying the two expansions:$(1 - 10x + 55x^2 - 90x^3 + \dots)(1 + 15x^2 + \dots)$The $x^3$ term is obtained by $(1 \cdot 0) + (-10x \cdot 15x^2) + (55x^2 \cdot 0) + (-90x^3 \cdot 1) = -150x^3 - 90x^3 = -240x^3$.Thus,the coefficient of $x^3$ is $-240$.

The coefficient of $x^3$ in the expansion of $(x^2 - x + 1)^{10} (x^2 + 1)^{15}$ is equal to:

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