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Question Number 42030 by Akashuac last updated on 17/Aug/18

$$\mathrm{x}+\frac{1}{\mathrm{x}}=5{\mathrm{x}}^5+\frac{1}{{\mathrm{x}}^5}=?$$

Commented by maxmathsup by imad last updated on 17/Aug/18

$$x+\frac{1}{x}=5\Rightarrow {(x+\frac{1}{x})}^5=5^5\Rightarrow \sum _{k=0}^5{C}_5^k{x}^k{\left(\frac{1}{x}\right)}^{5-k}=5^5\Rightarrow$$$$\sum _{k=0}^5{C}_5^k{x}^{2k-5}=5^5\Rightarrow x^{-5}+5x^{-3}+C_5^2{x}^{-1}+C_5^{3}x+5x^3+x^5=5^5\Rightarrow$$$$x^5+x^{-5}+5(x^3+x^{-3})+C_5^2(x+x^{-1})=5^5\Rightarrow$$$$x^5+x^{-5}=5^5-5(x^3+x^{-3})-C_5^2(x+\frac{1}{x})but$$$${(x+\frac{1}{x})}^3=5^3\Rightarrow x^3+3x^2\frac{1}{x}+3x\frac{1}{x^2}+\frac{1}{x^3}=5^3\Rightarrow$$$$x^3+x^{-3}+3(x+\frac{1}{x})=5^3\Rightarrow x^3+x^{-3}=5^3-3\times 5wehave{C}_5^2=\frac{5!}{2!3!}$$$$=\frac{5.4}{2}=10\Rightarrow x^5+x^{-5}=5^5-5(5^3-15)-10\times 5$$$$=5^5-5^4+5\times 15-50=5^4\times 4+25=4\times {25}^2+25=4{\left(\frac{100}{4}\right)}^2+25$$$$=\frac{{10}^4}{4}+25=\frac{10000}{4}+25=2500+25=2525.$$

Answered by MrW3 last updated on 17/Aug/18

$$x^5+\frac{1}{x^5}=$$$$=(x+\frac{1}{x})(x^4-x^3\times \frac{1}{x}+x^2\times \frac{1}{x^2}-x\times \frac{1}{x^3}+\frac{1}{x^4})$$$$=5(x^4-x^2+1-\frac{1}{x^2}+\frac{1}{x^4})$$$$=5(x^4+\frac{1}{x^4}+2x^2\times \frac{1}{x^2}-x^2-\frac{1}{x^2}-2x\times \frac{1}{x}+1-2+2)$$$$=5[{(x^2+\frac{1}{x^2})}^2-{(x+\frac{1}{x})}^2+1]$$$$=5[{(x^2+\frac{1}{x^2}+2x\times \frac{1}{x}-2)}^2-5^2+1]$$$$=5[{\{{(x+\frac{1}{x})}^2-2\}}^2-5^2+1]$$$$=5[{\{5^2-2\}}^2-5^2+1]$$$$=5[{23}^2-5^2+1]$$$$=5[28\times 18+1]$$$$=5\times 505$$$$=2525$$$$ingeneral:$$$$=a[{(a^2-2)}^2-a^2+1]$$$$=a[\{a(a+1)-2\}\{(a-1)a-2\}+1]$$

Commented by Akashuac last updated on 17/Aug/18

$$\mathrm{Thanks}\mathrm{you}\mathrm{Boss}$$

Answered by MJS last updated on 17/Aug/18

$$x+\frac{1}{x}=a$$$$x^2-ax+1=0$$$$x=\frac{a\pm \sqrt{a^2-4}}{2}$$$$\frac{1}{x}=\frac{2}{a\pm \sqrt{a^2-4}}=\frac{2(a\mp \sqrt{a^2-4})}{a^2-(a^2-4)}=\frac{a\mp \sqrt{a^2-4}}{2}$$$$\mathrm{we}\mathrm{can}\mathrm{write}x\mathrm{as}p\pm \sqrt{q}\mathrm{and}\frac{1}{x}\mathrm{as}p\mp \sqrt{q}$$$$p\in \mathbb{Q}\Rightarrow x^n+\frac{1}{x^n}=[{(p+\sqrt{q})}^n+{(p-\sqrt{q})}^n]\in \mathbb{Q}\mathrm{\forall }n\in \mathbb{Z}$$$$\mathrm{we}\mathrm{only}\mathrm{need}\mathrm{to}\mathrm{sum}\mathrm{up}\mathrm{the}{p}^i{\left(\sqrt{q}\right)}^j\mathrm{with}j=2k$$$$n=5\Rightarrow x^5+\frac{1}{x^5}=$$$$=2(p^5+10p^{3}q+5{pq}^2)=$$$$[p=\frac{a}{2}q=\frac{a^2-4}{4}]$$$$=a(a^4-5a^2+5)=$$$$[a=5]$$$$=2525$$

Commented by Akashuac last updated on 17/Aug/18

$$\mathrm{thanks}\mathrm{boss}$$

Answered by malwaan last updated on 17/Aug/18

$$\mathrm{by}\mathrm{using}$$$${(\mathrm{x}+\frac{1}{\mathrm{x}})}^5=5^5$$$${(\mathrm{x}+\frac{1}{\mathrm{x}})}^3=5^3\mathrm{we}\mathrm{find}$$$${\mathrm{x}}^5+\frac{1}{{\mathrm{x}}^5}=5^5-10\times 5-$$$$-5(5^3-3\times 5)$$$$=2525$$

Commented by Akashuac last updated on 17/Aug/18

$$\mathrm{thanks}\mathrm{boss}$$

Answered by math1967 last updated on 17/Aug/18

$${(x+\frac{1}{x})}^2=25$$$$x^2+\frac{1}{x^2}=23.........1$$$$x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=125$$$$x^3+\frac{1}{x^3}=125-15=110..........2$$$$\therefore (x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})=23\times 110$$$$x^5+\frac{1}{x^5}+x+\frac{1}{x}=2530$$$$\therefore x^5+\frac{1}{x^5}=2530-5=2525$$

Commented by Akashuac last updated on 17/Aug/18

$$\mathrm{thanks}\mathrm{boss}$$

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