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Question-200622

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Question Number 200622 by sonukgindia last updated on 21/Nov/23
Answered by Rasheed.Sindhi last updated on 21/Nov/23
•(a+b+c)^2 =4^2 =16 a^2 +b^2 +c^2 +2(ab+bc+ca)=16 ab+bc+ca=((16−10)/2)=3 •a^3 +b^3 +c^3 −3abc=(a+b+c)(a^2 +b^2 +c^2 −(ab+bc+ca)) 22−3abc=(4)(10−3)=28 abc=((22−28)/3)=−2 •(ab+bc+ca)^2 =3^2 =9 a^2 b^2 +b^2 c^2 +c^2 a^2 +2abc(a+b+c)=9 a^2 b^2 +b^2 c^2 +c^2 a^2 +2(−2)(4)=9 a^2 b^2 +b^2 c^2 +c^2 a^2 =9+16=25 •(a^2 +b^2 +c^2 )^2 =10^2 =100 a^4 +b^4 +c^4 +2(a^2 b^2 +b^2 c^2 +c^2 a^2 )=100 a^4 +b^4 +c^4 +2(25)=100 a^4 +b^4 +c^4 =100−50=50
$$\bullet\left({a}+{b}+{c}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} =\mathrm{16} \\ $$$$\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right)=\mathrm{16} \\ $$$$\:\:\:{ab}+{bc}+{ca}=\frac{\mathrm{16}−\mathrm{10}}{\mathrm{2}}=\mathrm{3} \\ $$$$\bullet{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\left({ab}+{bc}+{ca}\right)\right) \\ $$$$\:\:\:\mathrm{22}−\mathrm{3}{abc}=\left(\mathrm{4}\right)\left(\mathrm{10}−\mathrm{3}\right)=\mathrm{28} \\ $$$$\:\:\:{abc}=\frac{\mathrm{22}−\mathrm{28}}{\mathrm{3}}=−\mathrm{2} \\ $$$$\bullet\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$$$\:\:\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{2}{abc}\left({a}+{b}+{c}\right)=\mathrm{9} \\ $$$$\:\:\:\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} +\mathrm{2}\left(−\mathrm{2}\right)\left(\mathrm{4}\right)=\mathrm{9} \\ $$$$\:\:\:\:\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} =\mathrm{9}+\mathrm{16}=\mathrm{25} \\ $$$$\bullet\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} =\mathrm{100} \\ $$$$\:\:\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +\mathrm{2}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)=\mathrm{100} \\ $$$$\:\:\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +\mathrm{2}\left(\mathrm{25}\right)=\mathrm{100} \\ $$$$\:\:\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{100}−\mathrm{50}=\mathrm{50} \\ $$
Answered by mr W last updated on 21/Nov/23
p_1 =e_1 =4 p_2 =e_1 p_1 −2e_2 =4×4−2e_2 =10 ⇒e_2 =3 p_3 =e_1 p_2 −e_2 p_1 +3e_3 =4×10−3×4+3e_3 =22 ⇒e_3 =−2 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 =4×22−3×10−2×4=50 i.e. a^4 +b^4 +c^4 =50 ✓ generally we have p_n =4p_(n−1) −3p_(n−2) −2p_(n−3) p_n =a^n +b^n +c^n =2^n +(1+(√2))^n +(1−(√2))^n
$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{4} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} =\mathrm{4}×\mathrm{4}−\mathrm{2}{e}_{\mathrm{2}} =\mathrm{10}\:\Rightarrow{e}_{\mathrm{2}} =\mathrm{3} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} =\mathrm{4}×\mathrm{10}−\mathrm{3}×\mathrm{4}+\mathrm{3}{e}_{\mathrm{3}} =\mathrm{22}\:\Rightarrow{e}_{\mathrm{3}} =−\mathrm{2} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =\mathrm{4}×\mathrm{22}−\mathrm{3}×\mathrm{10}−\mathrm{2}×\mathrm{4}=\mathrm{50} \\ $$$${i}.{e}.\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{50}\:\checkmark \\ $$$$ \\ $$$${generally}\:{we}\:{have} \\ $$$${p}_{{n}} =\mathrm{4}{p}_{{n}−\mathrm{1}} −\mathrm{3}{p}_{{n}−\mathrm{2}} −\mathrm{2}{p}_{{n}−\mathrm{3}} \\ $$$${p}_{{n}} ={a}^{{n}} +{b}^{{n}} +{c}^{{n}} =\mathrm{2}^{{n}} +\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{{n}} +\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)^{{n}} \\ $$

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