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r-s-t-are-the-roots-of-x-3-5x-1-0-find-r-3-1-s-3-1-t-3-1-

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Question Number 223267 by behi834171 last updated on 19/Jul/25
r,s,t ; are the roots of: x^3 +5x+1=0 find: (r^3 −1)(s^3 −1)(t^3 −1)
$$\boldsymbol{{r}},\boldsymbol{{s}},\boldsymbol{{t}}\:;\:\boldsymbol{{are}}\:\boldsymbol{{the}}\:\boldsymbol{{roots}}\:\boldsymbol{{of}}: \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{5}\boldsymbol{{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{{find}}: \\ $$$$\:\:\:\:\:\:\left(\boldsymbol{{r}}^{\mathrm{3}} −\mathrm{1}\right)\left(\boldsymbol{{s}}^{\mathrm{3}} −\mathrm{1}\right)\left(\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{1}\right) \\ $$
Answered by gregori last updated on 20/Jul/25
x^3 −1 = −5x −2 r^3 −1 =−5r−2 s^3 −1= −5s−2 t^3 −1=−5t−2 (r^3 −1)(s^3 −1)(t^3 −1)= (−5r−2)(−5s−2)(−5t−2) = (25rs+10r+10s+4)(−5t−2) = −125rst−50rs−50rt−20r−50st−20s−20t−8 = −125−0−20−8 =−153
$$\:\:{x}^{\mathrm{3}} −\mathrm{1}\:=\:−\mathrm{5}{x}\:−\mathrm{2} \\ $$$$\:\:{r}^{\mathrm{3}} −\mathrm{1}\:=−\mathrm{5}{r}−\mathrm{2} \\ $$$$\:\:{s}^{\mathrm{3}} −\mathrm{1}=\:−\mathrm{5}{s}−\mathrm{2} \\ $$$$\:\:{t}^{\mathrm{3}} −\mathrm{1}=−\mathrm{5}{t}−\mathrm{2} \\ $$$$\:\:\left({r}^{\mathrm{3}} −\mathrm{1}\right)\left({s}^{\mathrm{3}} −\mathrm{1}\right)\left({t}^{\mathrm{3}} −\mathrm{1}\right)=\:\left(−\mathrm{5}{r}−\mathrm{2}\right)\left(−\mathrm{5}{s}−\mathrm{2}\right)\left(−\mathrm{5}{t}−\mathrm{2}\right) \\ $$$$\:=\:\left(\mathrm{25}{rs}+\mathrm{10}{r}+\mathrm{10}{s}+\mathrm{4}\right)\left(−\mathrm{5}{t}−\mathrm{2}\right) \\ $$$$\:\:=\:−\mathrm{125}{rst}−\mathrm{50}{rs}−\mathrm{50}{rt}−\mathrm{20}{r}−\mathrm{50}{st}−\mathrm{20}{s}−\mathrm{20}{t}−\mathrm{8} \\ $$$$\:\:=\:−\mathrm{125}−\mathrm{0}−\mathrm{20}−\mathrm{8} \\ $$$$\:\:=−\mathrm{153} \\ $$
Answered by mr W last updated on 20/Jul/25
x^3 −1=−(2+5x) (r^3 −1)(s^3 −1)(t^3 −1) =−(2+5r)(2+5s)(2+5t) =−[8+20(r+s+t)+50(rs+st+tr)+125rst] =−[8+20×0+50×5−125×1] =−133
$${x}^{\mathrm{3}} −\mathrm{1}=−\left(\mathrm{2}+\mathrm{5}{x}\right) \\ $$$$ \\ $$$$\left({r}^{\mathrm{3}} −\mathrm{1}\right)\left({s}^{\mathrm{3}} −\mathrm{1}\right)\left({t}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$=−\left(\mathrm{2}+\mathrm{5}{r}\right)\left(\mathrm{2}+\mathrm{5}{s}\right)\left(\mathrm{2}+\mathrm{5}{t}\right) \\ $$$$=−\left[\mathrm{8}+\mathrm{20}\left({r}+{s}+{t}\right)+\mathrm{50}\left({rs}+{st}+{tr}\right)+\mathrm{125}{rst}\right] \\ $$$$=−\left[\mathrm{8}+\mathrm{20}×\mathrm{0}+\mathrm{50}×\mathrm{5}−\mathrm{125}×\mathrm{1}\right] \\ $$$$=−\mathrm{133} \\ $$
Answered by behi834171 last updated on 22/Jul/25
x^3 +5x+1=(x−r)(x−s)(x−t) r^3 −1=(r−1)(r^2 +r+1)=(r−1)(r−𝛚)(r−𝛚^2 ) ⇒ { ((x=1⇒1^3 +5×1+1=(1−r)(1−s)(1−t))),((x=𝛚⇒𝛚^3 +5𝛚+1=(𝛚−r)(𝛚−s)(𝛚−t))),((x=𝛚^2 ⇒𝛚^6 +5𝛚^2 +1=(𝛚^2 −r)(𝛚^2 −s)(𝛚^2 −t))) :} ⇒Π(r^3 −1)=−(7)(2+5𝛚)(2+5𝛚^2 )= =−(7)[19+10(𝛚^2 +𝛚+1)]= =−7×19=−133. ■
$$\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{5}\boldsymbol{{x}}+\mathrm{1}=\left(\boldsymbol{{x}}−\boldsymbol{{r}}\right)\left(\boldsymbol{{x}}−\boldsymbol{{s}}\right)\left(\boldsymbol{{x}}−\boldsymbol{{t}}\right) \\ $$$$\boldsymbol{{r}}^{\mathrm{3}} −\mathrm{1}=\left(\boldsymbol{{r}}−\mathrm{1}\right)\left(\boldsymbol{{r}}^{\mathrm{2}} +\boldsymbol{{r}}+\mathrm{1}\right)=\left(\boldsymbol{{r}}−\mathrm{1}\right)\left(\boldsymbol{{r}}−\boldsymbol{\omega}\right)\left(\boldsymbol{{r}}−\boldsymbol{\omega}^{\mathrm{2}} \right) \\ $$$$\Rightarrow \\ $$$$\begin{cases}{\boldsymbol{{x}}=\mathrm{1}\Rightarrow\mathrm{1}^{\mathrm{3}} +\mathrm{5}×\mathrm{1}+\mathrm{1}=\left(\mathrm{1}−\boldsymbol{{r}}\right)\left(\mathrm{1}−\boldsymbol{{s}}\right)\left(\mathrm{1}−\boldsymbol{{t}}\right)}\\{\boldsymbol{{x}}=\boldsymbol{\omega}\Rightarrow\boldsymbol{\omega}^{\mathrm{3}} +\mathrm{5}\boldsymbol{\omega}+\mathrm{1}=\left(\boldsymbol{\omega}−\boldsymbol{{r}}\right)\left(\boldsymbol{\omega}−\boldsymbol{{s}}\right)\left(\boldsymbol{\omega}−\boldsymbol{{t}}\right)}\\{\boldsymbol{{x}}=\boldsymbol{\omega}^{\mathrm{2}} \Rightarrow\boldsymbol{\omega}^{\mathrm{6}} +\mathrm{5}\boldsymbol{\omega}^{\mathrm{2}} +\mathrm{1}=\left(\boldsymbol{\omega}^{\mathrm{2}} −\boldsymbol{{r}}\right)\left(\boldsymbol{\omega}^{\mathrm{2}} −\boldsymbol{{s}}\right)\left(\boldsymbol{\omega}^{\mathrm{2}} −\boldsymbol{{t}}\right)}\end{cases} \\ $$$$\Rightarrow\Pi\left(\boldsymbol{{r}}^{\mathrm{3}} −\mathrm{1}\right)=−\left(\mathrm{7}\right)\left(\mathrm{2}+\mathrm{5}\boldsymbol{\omega}\right)\left(\mathrm{2}+\mathrm{5}\boldsymbol{\omega}^{\mathrm{2}} \right)= \\ $$$$=−\left(\mathrm{7}\right)\left[\mathrm{19}+\mathrm{10}\left(\boldsymbol{\omega}^{\mathrm{2}} +\boldsymbol{\omega}+\mathrm{1}\right)\right]= \\ $$$$=−\mathrm{7}×\mathrm{19}=−\mathrm{133}.\:\:\blacksquare \\ $$

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