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If-B-R-P-1-B-2-R-2-P-2-17-B-3-R-3-P-3-11-then-B-5-R-5-P-5-

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Question Number 186929 by cortano12 last updated on 12/Feb/23
If { ((B+R+P=−1)),((B^2 +R^2 +P^2 =17)),((B^3 +R^3 +P^3 =11)) :} then B^5 +R^5 +P^5 =?
$$\:\:{If}\:\begin{cases}{{B}+{R}+{P}=−\mathrm{1}}\\{{B}^{\mathrm{2}} +{R}^{\mathrm{2}} +{P}^{\mathrm{2}} =\mathrm{17}}\\{{B}^{\mathrm{3}} +{R}^{\mathrm{3}} +{P}^{\mathrm{3}} =\mathrm{11}}\end{cases} \\ $$$$\:{then}\:{B}^{\mathrm{5}} +{R}^{\mathrm{5}} +{P}^{\mathrm{5}} \:=? \\ $$
Answered by mr W last updated on 12/Feb/23
method I p_1 =e_1 =−1 p_2 =e_1 p_1 −2e_2 17=(−1)^2 −2e_2 ⇒e_2 =−8 p_3 =e_1 p_2 −e_2 p_1 +3e_3 11=−1×17−(−8)(−1)+3e_3 ⇒e_3 =12 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 =−1×11+8×17−12×1=113 p_5 =e_1 p_4 −e_2 p_3 +e_3 p_2 =−1×113+8×11+12×17=179 i.e. B^5 +R^5 +P^5 =179
$${method}\:{I} \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =−\mathrm{1} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} \\ $$$$\mathrm{17}=\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \:\Rightarrow{e}_{\mathrm{2}} =−\mathrm{8} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\mathrm{11}=−\mathrm{1}×\mathrm{17}−\left(−\mathrm{8}\right)\left(−\mathrm{1}\right)+\mathrm{3}{e}_{\mathrm{3}} \:\Rightarrow{e}_{\mathrm{3}} =\mathrm{12} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =−\mathrm{1}×\mathrm{11}+\mathrm{8}×\mathrm{17}−\mathrm{12}×\mathrm{1}=\mathrm{113} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} =−\mathrm{1}×\mathrm{113}+\mathrm{8}×\mathrm{11}+\mathrm{12}×\mathrm{17}=\mathrm{179} \\ $$$${i}.{e}.\:{B}^{\mathrm{5}} +{R}^{\mathrm{5}} +{P}^{\mathrm{5}} =\mathrm{179} \\ $$
Answered by mr W last updated on 12/Feb/23
method II i write a,b,c instead of B,R,P. a+b+c=−1 (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ca) (−1)^2 =17+2(ab+bc+ca) ⇒ab+bc+ca=−8 (a+b+c)^3 =a^3 +b^3 +c^3 −3abc+3(a+b+c)(ab+bc+ca) (−1)^3 =11−3abc+3(−1)(−8) ⇒abc=12 (ab+bc+ca)^2 =a^2 b^2 +b^2 c^2 +c^2 a^2 +2abc(a+b+c) (−8)^2 =a^2 b^2 +b^2 c^2 +c^2 a^2 +2×12(−1) ⇒a^2 b^2 +b^2 c^2 +c^2 a^2 =88 (a^2 +b^2 +c^2 )(a^3 +b^3 +c^3 )=a^5 +b^5 +c^5 +(a+b+c)(a^2 b^2 +b^2 c^2 +c^2 a^2 )−abc(ab+bc+ca) (17)(11)=a^5 +b^5 +c^5 +(−1)(88)−12(−8) ⇒a^5 +b^5 +c^5 =179
$${method}\:{II} \\ $$$${i}\:{write}\:{a},{b},{c}\:{instead}\:{of}\:{B},{R},{P}. \\ $$$${a}+{b}+{c}=−\mathrm{1} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{17}+\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\Rightarrow{ab}+{bc}+{ca}=−\mathrm{8} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}+\mathrm{3}\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right) \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{11}−\mathrm{3}{abc}+\mathrm{3}\left(−\mathrm{1}\right)\left(−\mathrm{8}\right) \\ $$$$\Rightarrow{abc}=\mathrm{12} \\ $$$$\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{2}{abc}\left({a}+{b}+{c}\right) \\ $$$$\left(−\mathrm{8}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{2}×\mathrm{12}\left(−\mathrm{1}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} =\mathrm{88} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)={a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} +\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)−{abc}\left({ab}+{bc}+{ca}\right) \\ $$$$\left(\mathrm{17}\right)\left(\mathrm{11}\right)={a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} +\left(−\mathrm{1}\right)\left(\mathrm{88}\right)−\mathrm{12}\left(−\mathrm{8}\right) \\ $$$$\Rightarrow{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} =\mathrm{179} \\ $$
Answered by horsebrand11 last updated on 12/Feb/23
T_n =B^n +R^n +P^n T_0 =B^0 +R^0 +P^0 =3 T_1 =−1=α_1 T_2 =B^2 +R^2 +P^2 =(B+R+P)^2 −2(BR+BP+RP) ⇒17=1−2(BR+BP+RP) ⇒ BR+BP+RP = −8=α_2 T_n = α_1 T_(n−1) −α_2 T_(n−2) +α_3 T_(n−3) T_3 =(−1).17−(−8)(−1)+α_3 (3)=11 ⇒α_3 = 12 T_4 =α_1 T_3 −α_2 T_2 +α_3 T_1 ⇒T_4 = (−1)(11)−(−8)(17)+(12).(−1) ⇒T_4 = 113 T_5 =α_1 T_4 −α_2 T_3 +α_3 T_2 ⇒T_5 =(−1)(113)−(−8)(11)+(12)(17) ⇒T_5 =179
$$\:{T}_{{n}} ={B}^{{n}} +{R}^{{n}} +{P}^{{n}} \: \\ $$$${T}_{\mathrm{0}} ={B}^{\mathrm{0}} +{R}^{\mathrm{0}} +{P}^{\mathrm{0}} =\mathrm{3} \\ $$$$\:{T}_{\mathrm{1}} =−\mathrm{1}=\alpha_{\mathrm{1}} \\ $$$${T}_{\mathrm{2}} ={B}^{\mathrm{2}} +{R}^{\mathrm{2}} +{P}^{\mathrm{2}} =\left({B}+{R}+{P}\right)^{\mathrm{2}} −\mathrm{2}\left({BR}+{BP}+{RP}\right) \\ $$$$\:\Rightarrow\mathrm{17}=\mathrm{1}−\mathrm{2}\left({BR}+{BP}+{RP}\right) \\ $$$$\:\Rightarrow\:{BR}+{BP}+{RP}\:=\:−\mathrm{8}=\alpha_{\mathrm{2}} \\ $$$${T}_{{n}} =\:\alpha_{\mathrm{1}} {T}_{{n}−\mathrm{1}} −\alpha_{\mathrm{2}} {T}_{{n}−\mathrm{2}} +\alpha_{\mathrm{3}} {T}_{{n}−\mathrm{3}} \\ $$$${T}_{\mathrm{3}} =\left(−\mathrm{1}\right).\mathrm{17}−\left(−\mathrm{8}\right)\left(−\mathrm{1}\right)+\alpha_{\mathrm{3}} \left(\mathrm{3}\right)=\mathrm{11} \\ $$$$\:\Rightarrow\alpha_{\mathrm{3}} =\:\mathrm{12} \\ $$$${T}_{\mathrm{4}} =\alpha_{\mathrm{1}} {T}_{\mathrm{3}} −\alpha_{\mathrm{2}} {T}_{\mathrm{2}} +\alpha_{\mathrm{3}} {T}_{\mathrm{1}} \\ $$$$\Rightarrow{T}_{\mathrm{4}} =\:\left(−\mathrm{1}\right)\left(\mathrm{11}\right)−\left(−\mathrm{8}\right)\left(\mathrm{17}\right)+\left(\mathrm{12}\right).\left(−\mathrm{1}\right) \\ $$$$\Rightarrow{T}_{\mathrm{4}} =\:\mathrm{113} \\ $$$${T}_{\mathrm{5}} =\alpha_{\mathrm{1}} {T}_{\mathrm{4}} −\alpha_{\mathrm{2}} {T}_{\mathrm{3}} +\alpha_{\mathrm{3}} {T}_{\mathrm{2}} \\ $$$$\Rightarrow{T}_{\mathrm{5}} =\left(−\mathrm{1}\right)\left(\mathrm{113}\right)−\left(−\mathrm{8}\right)\left(\mathrm{11}\right)+\left(\mathrm{12}\right)\left(\mathrm{17}\right) \\ $$$$\Rightarrow{T}_{\mathrm{5}} =\mathrm{179} \\ $$

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