[tngngoapm]

$Problem1......(x^{3}+1)^{2560} หารด้วย x^{2}+x+1 เหลือเศษเท่าไหร่?$
วิธีทำ
1. ให้ $P(x)=(x^{3}+1)^{2560}$.........และหารากของสมการ $x^{2}+x+1=0$
$z_{1}=-\frac{1}{2}+\frac{\sqrt{3} }{2}i=cos\frac{2\pi }{3}+isin\frac{2\pi }{3} $
$z_{2}=-\frac{1}{2}-\frac{\sqrt{3} }{2}i=cos\frac{4\pi }{3}+isin\frac{4\pi }{3} $
2.เศษจะอยู่ในรูปแบบ $px+q,p,q\in R$.....โดยที่
$p=\frac{P(z_{1})-P(z_{2})}{z_{1}-z_{2}}$
$q=\frac{z_{1}P(z_{2})-z_{2}P(z_{1})}{z_{1}-z_{2}} $
......หา $P(z_{1})=([cos\frac{2\pi }{3}+isin\frac{2\pi }{3}]^{3}+1)^{2560}=(1+1)^{2560}=2^{2560}$
$P(z_{2})=([cos\frac{4\pi }{3}+isin\frac{4\pi }{3}]^{3}+1)^{2560}=(1+1)^{2560}=2^{2560}$
$\therefore p=\frac{2^{2560}-2^{2560}}{(-\frac{1}{2}+\frac{\sqrt{3} }{2}i)-(-\frac{1}{2}-\frac{\sqrt{3} }{2}i)}=0$
$\therefore q=\frac{z_{1}(2^{2560})-z_{2}(2^{2560})}{z_{1}-z_{2}}=\frac{(2^{2560})(z_{1}-z_{2})}{z_{1}-z_{2}}=2^{2560} $
3.เศษคือ $2^{2560}$...........$Ans$