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Question Number 59518 by Mr X pcx last updated on 11/May/19

$$ifx+y+z=1$$$$x^2+y^2+z^2=2$$$$x^3+y^3+z^3=3calculste$$$$x^5+y^5+z^5$$

Commented by MJS last updated on 11/May/19

$$\mathrm{we}\mathrm{had}\mathrm{this}\mathrm{several}\mathrm{times}\mathrm{before}.\mathrm{the}\mathrm{answer}\mathrm{is}6$$

Answered by MJS last updated on 11/May/19

$$\mathrm{put}x=a;y=b-c\mathrm{i};z=b+c\mathrm{i}$$$$x+y+z=1\Rightarrow a+2b=1\Rightarrow a=1-2b$$$$x^2+y^2+z^2=2\Rightarrow a^2+2b^2-2c^2=2\Rightarrow$$$$\Rightarrow 6b^2-4b-2c^2+1=2\Rightarrow c=\frac{\sqrt{2(6b^2-4b-1)}}{2}$$$$x^3+y^3+z^3=3\Rightarrow a^3+2b(b^2-3c^2)=3\Rightarrow$$$$\Rightarrow -24b^3+24b^2-3b+1=3\Rightarrow$$$$\Rightarrow b^3-b^2+\frac{1}{8}b+\frac{1}{12}=0$$$$$$$$x^5+y^5+z^5=a^5+2b(b^4-10b^2{c}^2+5c^4)=$$$$=-60b^3+60b^2-\frac{15}{2}b+1=-60(b^3-b^2+\frac{1}{8}b-\frac{1}{60})$$$$$$$$b^3-b^2+\frac{1}{8}b+\frac{1}{12}=0\Rightarrow b^3-b^2+\frac{1}{8}b-\frac{1}{60}=-\frac{1}{10}\Rightarrow$$$$\Rightarrow -60(b^3-b^2+\frac{1}{8}b-\frac{1}{60})=6\Rightarrow$$$$\Rightarrow x^5+y^5+z^5=6$$

Commented by MJS last updated on 11/May/19

$$x+y+z=\alpha$$$$x^2+y^2+z^2=\beta$$$$x^3+y^3+z^3=\gamma$$$$\Rightarrow$$$$x^4+y^4+z^4=\frac{1}{6}{\alpha }^4-{\alpha }^2\beta +\frac{4}{3}\alpha \gamma +\frac{1}{2}{\beta }^2$$$$x^5+y^5+z^5=\frac{1}{6}{\alpha }^5-\frac{5}{6}{\alpha }^3\beta +\frac{5}{6}{\alpha }^2\gamma +\frac{5}{6}\beta \gamma$$

Commented by maxmathsup by imad last updated on 11/May/19

$$thankssirmjs$$

Commented by Tawa1 last updated on 21/Jul/19

$$\mathrm{Sir}\mathrm{reference}\mathrm{to}\mathrm{the}\mathrm{identity}.$$$$\mathrm{I}\mathrm{did}\mathrm{not}\mathrm{get}6\mathrm{when}\mathrm{i}\mathrm{substitute}\alpha =1,\beta =2,\gamma =3\mathrm{for}{\alpha }^5+{\beta }^5+{\gamma }^5$$$$\mathrm{and}\mathrm{i}\mathrm{got}\frac{25}{6}\mathrm{for}{\alpha }^4+{\beta }^4+{\gamma }^4$$

Commented by MJS last updated on 21/Jul/19

$$\frac{25}{6}\mathrm{is}\mathrm{the}\mathrm{right}\mathrm{answer}$$$${\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{look}\mathrm{into}\mathrm{it}\mathrm{again}$$

Commented by Tawa1 last updated on 21/Jul/19

$$\mathrm{Okay}\mathrm{sir},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$$$\mathrm{That}\mathrm{means}\mathrm{the}\mathrm{identity}\mathrm{is}\mathrm{correct}\mathrm{for}{\alpha }^4+{\beta }^4+{\gamma }^4=\frac{25}{4}$$

Commented by MJS last updated on 21/Jul/19

$$\mathrm{yes}$$

Commented by MJS last updated on 21/Jul/19

$$\frac{1}{6}\times 1^5-\frac{5}{6}\times 1^3\times 2+\frac{5}{6}\times 1^2\times 3+\frac{5}{6}\times 2\times 3=$$$$=\frac{1}{6}-\frac{10}{6}+\frac{15}{6}+\frac{30}{6}=\frac{36}{6}=6$$$${\mathrm{you}}^{\prime }\mathrm{ll}\mathrm{have}\mathrm{to}\mathrm{check}\mathrm{your}\mathrm{calculation}\mathrm{for}\mathrm{typos}$$

Commented by Tawa1 last updated on 21/Jul/19

$$\mathrm{Wow}\mathrm{great}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by Tawa1 last updated on 21/Jul/19

$$\mathrm{One}\mathrm{more}\mathrm{thing}\mathrm{sir}.\mathrm{How}\mathrm{do}\mathrm{you}\mathrm{get}\mathrm{the}\mathrm{identity}\mathrm{sir}.$$$$\mathrm{when}\mathrm{you}\mathrm{are}\mathrm{less}\mathrm{busy}$$

Commented by MJS last updated on 21/Jul/19

$${\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{repost}\mathrm{this}\mathrm{as}\mathrm{a}\mathrm{new}\mathrm{question}\mathrm{plus}\mathrm{answer}$$$$\mathrm{plus}\mathrm{explanation}$$

Commented by Tawa1 last updated on 21/Jul/19

$$\mathrm{Wow}\mathrm{great}.\mathrm{I}\mathrm{will}\mathrm{learn}\mathrm{from}\mathrm{it}$$

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