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Question Number 59518 by Mr X pcx last updated on 11/May/19
if x+y+z=1 x^2 +y^2 +z^2 =2 x^3 +y^3 +z^3 =3 calculste x^5 +y^5 +z^5
$${if}\:\:{x}+{y}+{z}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:+{z}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{\mathrm{3}} \:+{y}^{\mathrm{3}} \:+{z}^{\mathrm{3}} =\mathrm{3}\:\:{calculste} \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} \:+{z}^{\mathrm{5}} \\ $$
Commented by MJS last updated on 11/May/19
we had this several times before. the answer is 6
$$\mathrm{we}\:\mathrm{had}\:\mathrm{this}\:\mathrm{several}\:\mathrm{times}\:\mathrm{before}.\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{6} \\ $$
Answered by MJS last updated on 11/May/19
put x=a; y=b−ci; z=b+ci x+y+z=1 ⇒ a+2b=1 ⇒ a=1−2b x^2 +y^2 +z^2 =2 ⇒ a^2 +2b^2 −2c^2 =2 ⇒ ⇒ 6b^2 −4b−2c^2 +1=2 ⇒ c=((√(2(6b^2 −4b−1)))/2) x^3 +y^3 +z^3 =3 ⇒ a^3 +2b(b^2 −3c^2 )=3 ⇒ ⇒ −24b^3 +24b^2 −3b+1=3 ⇒ ⇒ b^3 −b^2 +(1/8)b+(1/(12))=0 x^5 +y^5 +z^5 =a^5 +2b(b^4 −10b^2 c^2 +5c^4 )= =−60b^3 +60b^2 −((15)/2)b+1=−60(b^3 −b^2 +(1/8)b−(1/(60))) b^3 −b^2 +(1/8)b+(1/(12))=0 ⇒ b^3 −b^2 +(1/8)b−(1/(60))=−(1/(10)) ⇒ ⇒ −60(b^3 −b^2 +(1/8)b−(1/(60)))=6 ⇒ ⇒ x^5 +y^5 +z^5 =6
$$\mathrm{put}\:{x}={a};\:{y}={b}−{c}\mathrm{i};\:{z}={b}+{c}\mathrm{i} \\ $$$${x}+{y}+{z}=\mathrm{1}\:\Rightarrow\:{a}+\mathrm{2}{b}=\mathrm{1}\:\Rightarrow\:{a}=\mathrm{1}−\mathrm{2}{b} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{2}\:\Rightarrow\:{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} =\mathrm{2}\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{6}{b}^{\mathrm{2}} −\mathrm{4}{b}−\mathrm{2}{c}^{\mathrm{2}} +\mathrm{1}=\mathrm{2}\:\Rightarrow\:{c}=\frac{\sqrt{\mathrm{2}\left(\mathrm{6}{b}^{\mathrm{2}} −\mathrm{4}{b}−\mathrm{1}\right)}}{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{3}\:\Rightarrow\:{a}^{\mathrm{3}} +\mathrm{2}{b}\left({b}^{\mathrm{2}} −\mathrm{3}{c}^{\mathrm{2}} \right)=\mathrm{3}\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:−\mathrm{24}{b}^{\mathrm{3}} +\mathrm{24}{b}^{\mathrm{2}} −\mathrm{3}{b}+\mathrm{1}=\mathrm{3}\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:{b}^{\mathrm{3}} −{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}{b}+\frac{\mathrm{1}}{\mathrm{12}}=\mathrm{0} \\ $$$$ \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} ={a}^{\mathrm{5}} +\mathrm{2}{b}\left({b}^{\mathrm{4}} −\mathrm{10}{b}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{5}{c}^{\mathrm{4}} \right)= \\ $$$$\:\:\:\:\:=−\mathrm{60}{b}^{\mathrm{3}} +\mathrm{60}{b}^{\mathrm{2}} −\frac{\mathrm{15}}{\mathrm{2}}{b}+\mathrm{1}=−\mathrm{60}\left({b}^{\mathrm{3}} −{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}{b}−\frac{\mathrm{1}}{\mathrm{60}}\right) \\ $$$$ \\ $$$${b}^{\mathrm{3}} −{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}{b}+\frac{\mathrm{1}}{\mathrm{12}}=\mathrm{0}\:\Rightarrow\:{b}^{\mathrm{3}} −{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}{b}−\frac{\mathrm{1}}{\mathrm{60}}=−\frac{\mathrm{1}}{\mathrm{10}}\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:−\mathrm{60}\left({b}^{\mathrm{3}} −{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{8}}{b}−\frac{\mathrm{1}}{\mathrm{60}}\right)=\mathrm{6}\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:{x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} =\mathrm{6} \\ $$
Commented by MJS last updated on 11/May/19
x+y+z=α x^2 +y^2 +z^2 =β x^3 +y^3 +z^3 =γ ⇒ x^4 +y^4 +z^4 =(1/6)α^4 −α^2 β+(4/3)αγ+(1/2)β^2 x^5 +y^5 +z^5 =(1/6)α^5 −(5/6)α^3 β+(5/6)α^2 γ+(5/6)βγ
$${x}+{y}+{z}=\alpha \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\beta \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\gamma \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{6}}\alpha^{\mathrm{4}} −\alpha^{\mathrm{2}} \beta+\frac{\mathrm{4}}{\mathrm{3}}\alpha\gamma+\frac{\mathrm{1}}{\mathrm{2}}\beta^{\mathrm{2}} \\ $$$${x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{6}}\alpha^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{6}}\alpha^{\mathrm{3}} \beta+\frac{\mathrm{5}}{\mathrm{6}}\alpha^{\mathrm{2}} \gamma+\frac{\mathrm{5}}{\mathrm{6}}\beta\gamma \\ $$
Commented by maxmathsup by imad last updated on 11/May/19
thanks sir mjs
$${thanks}\:{sir}\:{mjs} \\ $$
Commented by Tawa1 last updated on 21/Jul/19
Sir reference to the identity. I did not get 6 when i substitute α = 1, β = 2, γ = 3 for α^5 + β^5 + γ^5 and i got ((25)/6) for α^4 + β^4 + γ^4
$$\mathrm{Sir}\:\mathrm{reference}\:\mathrm{to}\:\mathrm{the}\:\mathrm{identity}. \\ $$$$\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{get}\:\mathrm{6}\:\mathrm{when}\:\mathrm{i}\:\mathrm{substitute}\:\:\alpha\:=\:\mathrm{1},\:\beta\:=\:\mathrm{2},\:\gamma\:=\:\mathrm{3}\:\:\:\:\mathrm{for}\:\:\:\alpha^{\mathrm{5}} \:+\:\beta^{\mathrm{5}} \:+\:\gamma^{\mathrm{5}} \\ $$$$\mathrm{and}\:\mathrm{i}\:\mathrm{got}\:\:\frac{\mathrm{25}}{\mathrm{6}}\:\:\mathrm{for}\:\:\:\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:+\:\gamma^{\mathrm{4}} \\ $$
Commented by MJS last updated on 21/Jul/19
((25)/6) is the right answer I′ll look into it again
$$\frac{\mathrm{25}}{\mathrm{6}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{right}\:\mathrm{answer} \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{look}\:\mathrm{into}\:\mathrm{it}\:\mathrm{again} \\ $$
Commented by Tawa1 last updated on 21/Jul/19
Okay sir, God bless you sir. That means the identity is correct for α^4 + β^4 + γ^4 = ((25)/4)
$$\mathrm{Okay}\:\mathrm{sir},\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\:\:\:\:\mathrm{That}\:\mathrm{means}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{for}\:\:\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:+\:\gamma^{\mathrm{4}} \:=\:\frac{\mathrm{25}}{\mathrm{4}} \\ $$
Commented by MJS last updated on 21/Jul/19
yes
$$\mathrm{yes} \\ $$
Commented by MJS last updated on 21/Jul/19
(1/6)×1^5 −(5/6)×1^3 ×2+(5/6)×1^2 ×3+(5/6)×2×3= =(1/6)−((10)/6)+((15)/6)+((30)/6)=((36)/6)=6 you′ll have to check your calculation for typos
$$\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{1}^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{6}}×\mathrm{1}^{\mathrm{3}} ×\mathrm{2}+\frac{\mathrm{5}}{\mathrm{6}}×\mathrm{1}^{\mathrm{2}} ×\mathrm{3}+\frac{\mathrm{5}}{\mathrm{6}}×\mathrm{2}×\mathrm{3}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{10}}{\mathrm{6}}+\frac{\mathrm{15}}{\mathrm{6}}+\frac{\mathrm{30}}{\mathrm{6}}=\frac{\mathrm{36}}{\mathrm{6}}=\mathrm{6} \\ $$$$\mathrm{you}'\mathrm{ll}\:\mathrm{have}\:\mathrm{to}\:\mathrm{check}\:\mathrm{your}\:\mathrm{calculation}\:\mathrm{for}\:\mathrm{typos} \\ $$
Commented by Tawa1 last updated on 21/Jul/19
Wow great sir. God bless you sir.
$$\mathrm{Wow}\:\mathrm{great}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by Tawa1 last updated on 21/Jul/19
One more thing sir. How do you get the identity sir. when you are less busy
$$\mathrm{One}\:\mathrm{more}\:\mathrm{thing}\:\mathrm{sir}.\:\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{sir}. \\ $$$$\mathrm{when}\:\mathrm{you}\:\mathrm{are}\:\mathrm{less}\:\mathrm{busy} \\ $$
Commented by MJS last updated on 21/Jul/19
I′ll repost this as a new question plus answer plus explanation
$$\mathrm{I}'\mathrm{ll}\:\mathrm{repost}\:\mathrm{this}\:\mathrm{as}\:\mathrm{a}\:\mathrm{new}\:\mathrm{question}\:\mathrm{plus}\:\mathrm{answer} \\ $$$$\mathrm{plus}\:\mathrm{explanation} \\ $$
Commented by Tawa1 last updated on 21/Jul/19
Wow great. I will learn from it
$$\mathrm{Wow}\:\mathrm{great}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{learn}\:\mathrm{from}\:\mathrm{it} \\ $$

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