Question Number 223386 by behi834171 last updated on 23/Jul/25
Answered by mr W last updated on 23/Jul/25
![still using my method generally for x^n +kx+p=0 x^n −a=−k(x+((p+a)/k))=−ky (x+((p+a)/k)−((p+a)/k))^n +k(x+((p+a)/k)−((p+a)/k))+p=0 ⇒(y−((p+a)/k))^n +k(y−((p+a)/k))+p=0 ⇒...+...y^2 +...y+(−((p+a)/k))^n −1=0 (only constant term) Π_(m=1) ^n y_m =(−1)^n [(−((p+a)/k))^n −1]=(((p+a)/k))^n −(−1)^n Π_(m=1) ^n (x_m ^n −a)=(−k)^n Π_(m=1) ^n y_m =(−k)^n [(((p+a)/k))^n −(−1)^n ] =(−1)^n (p+a)^n −k^n examples: n=4, k=4, p=1, a=1 ⇒Π(x^4 −1)=(−1)^4 ×(1+1)^4 −4^4 =−240 n=3, k=5, p=1, a=1 ⇒Π(x^3 −1)=(−1)^3 ×(1+1)^3 −5^3 =−133](https://www.tinkutara.com/question/Q223388.png)
$${still}\:{using}\:{my}\:{method}\:{generally}\:{for} \\ $$$$\boldsymbol{{x}}^{\boldsymbol{{n}}} +\boldsymbol{{kx}}+\boldsymbol{{p}}=\mathrm{0} \\ $$$${x}^{{n}} −{a}=−{k}\left({x}+\frac{{p}+{a}}{{k}}\right)=−{ky} \\ $$$$\left({x}+\frac{{p}+{a}}{{k}}−\frac{{p}+{a}}{{k}}\right)^{{n}} +{k}\left({x}+\frac{{p}+{a}}{{k}}−\frac{{p}+{a}}{{k}}\right)+{p}=\mathrm{0} \\ $$$$\Rightarrow\left({y}−\frac{{p}+{a}}{{k}}\right)^{{n}} +{k}\left({y}−\frac{{p}+{a}}{{k}}\right)+{p}=\mathrm{0} \\ $$$$\Rightarrow…+…{y}^{\mathrm{2}} +…{y}+\left(−\frac{{p}+{a}}{{k}}\right)^{{n}} −\mathrm{1}=\mathrm{0}\:\:\:\left({only}\:{constant}\:{term}\right) \\ $$$$\underset{{m}=\mathrm{1}} {\overset{{n}} {\prod}}{y}_{{m}} =\left(−\mathrm{1}\right)^{{n}} \left[\left(−\frac{{p}+{a}}{{k}}\right)^{{n}} −\mathrm{1}\right]=\left(\frac{{p}+{a}}{{k}}\right)^{{n}} −\left(−\mathrm{1}\right)^{{n}} \\ $$$$\underset{{m}=\mathrm{1}} {\overset{{n}} {\prod}}\left({x}_{{m}} ^{{n}} −{a}\right)=\left(−{k}\right)^{{n}} \underset{{m}=\mathrm{1}} {\overset{{n}} {\prod}}{y}_{{m}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(−{k}\right)^{{n}} \left[\left(\frac{{p}+{a}}{{k}}\right)^{{n}} −\left(−\mathrm{1}\right)^{{n}} \right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(−\mathrm{1}\right)^{{n}} \left({p}+{a}\right)^{{n}} −{k}^{{n}} \\ $$$${examples}: \\ $$$${n}=\mathrm{4},\:{k}=\mathrm{4},\:{p}=\mathrm{1},\:{a}=\mathrm{1} \\ $$$$\Rightarrow\Pi\left({x}^{\mathrm{4}} −\mathrm{1}\right)=\left(−\mathrm{1}\right)^{\mathrm{4}} ×\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{4}} −\mathrm{4}^{\mathrm{4}} =−\mathrm{240} \\ $$$${n}=\mathrm{3},\:{k}=\mathrm{5},\:{p}=\mathrm{1},\:{a}=\mathrm{1} \\ $$$$\Rightarrow\Pi\left({x}^{\mathrm{3}} −\mathrm{1}\right)=\left(−\mathrm{1}\right)^{\mathrm{3}} ×\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{5}^{\mathrm{3}} =−\mathrm{133} \\ $$
Commented by behi834171 last updated on 23/Jul/25
$${perfect}! \\ $$$${so}\:{beautiful}\:{and}\:{effective}\:{method}. \\ $$$${thank}\:{you}\:{very}\:{much}\:{dear}\:{master}, \\ $$$${for}\:{your}\:{time}\:{and}\:{favor}. \\ $$