The coefficient of x3 in the expansion of (1−x+x2)6 is

Solution:

$\begin{aligned} {(1 - x + x^{2})}^{6} = [1 - x (1 - x)]^{6} \\ =^{6} C_{0} -^{6} C_{1} x (1 - x) +^{6} C_{2} x^{2} (1 - x)^{2} -^{6} C_{3} x^{3} (1 - x)^{3} + \dots to 7 terms \\ =^{6} C_{0} -^{6} C_{1} x (1 - x) +^{6} C_{2} x^{2} (1 - 2 x + x^{2}) -^{6} C_{3} x^{3} (1 - 3 x + 3 x^{2} - x^{3}) + \dots to 7 terms \end{aligned}$

$∴$ Coefficient of $x^{3} = - 2 \cdot^{6} C_{2} -^{6} C_{3}$ (collecting coefficients of x3 from each term)

$= - 2 \frac{6!}{2! 4!} - \frac{6!}{3! 3!} = - 50$

The coefficient of x³ in the expansion of ${(1 - x + x^{2})}^{6}$ is

Solution:

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