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Question-152980

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Question Number 152980 by mr W last updated on 03/Sep/21
Commented by mathdanisur last updated on 03/Sep/21
a^7 +b^7 +c^7 = 7abc(ab+bc+ca)^2 a^4 +b^4 +c^4 = 2(ab+bc+ca)^2 S = ((7abc(ab+bc+ca)^2 )/(abc[2(ab+bc+ca)^2 ])) = (7/2) = 3,5
$$\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} +\mathrm{c}^{\mathrm{7}} \:=\:\mathrm{7abc}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)^{\mathrm{2}} \\ $$$$\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} +\mathrm{c}^{\mathrm{4}} \:=\:\mathrm{2}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)^{\mathrm{2}} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{7abc}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)^{\mathrm{2}} }{\mathrm{abc}\left[\mathrm{2}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)^{\mathrm{2}} \right]}\:=\:\frac{\mathrm{7}}{\mathrm{2}}\:=\:\mathrm{3},\mathrm{5} \\ $$
Commented by mr W last updated on 03/Sep/21
solution please!
$${solution}\:{please}! \\ $$
Commented by mr W last updated on 03/Sep/21
find (((a^2 +b^2 +c^2 )(a^5 +b^5 +c^5 ))/(a^7 +b^7 +c^7 ))=?
$${find}\:\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} \right)}{{a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} }=? \\ $$
Commented by mathdanisur last updated on 03/Sep/21
= ((10)/7)
$$=\:\frac{\mathrm{10}}{\mathrm{7}} \\ $$
Commented by mathdanisur last updated on 03/Sep/21
Thanks ser ((a^7 +b^7 +c^7 )/7) = ((a^5 +b^5 +c^5 )/5) ((a^2 +b^2 +c^2 )/2)
$$\mathrm{Thanks}\:\mathrm{ser} \\ $$$$\frac{\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} +\mathrm{c}^{\mathrm{7}} }{\mathrm{7}}\:=\:\frac{\mathrm{a}^{\mathrm{5}} +\mathrm{b}^{\mathrm{5}} +\mathrm{c}^{\mathrm{5}} }{\mathrm{5}}\:\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by mathdanisur last updated on 03/Sep/21
a+b+c=0 a^k +b^k +c^k =S_k ; k∈Z^+ ⇒ S_(k+3) = abc S_k + (1/2) S_2 S_(k+1) a+b+c=0 ; a;b;c∈IR ⇒ S_0 =3 ⇒ a^0 +b^0 +c^0 =3 ⇒ S_1 =0 ⇒ a+b+c=0 ⇒ S_2 ^2 =2S_4 ⇒ (a^2 +b^2 +c^2 )^2 =2(a^4 +b^4 +c^4 ) ⇒ S_3 =3abc ⇒ a^3 +b^3 +c^3 =3abc ⇒ 6S_5 =5S_2 S_3 ⇒ 6(a^5 +b^5 +c^5 )=5(a^2 +b^2 +c^2 )(a^3 +b^3 +c^3 ) ⇒ 6S_7 =7S_3 S_4 ⇒ 6(a^7 +b^7 +c^7 )=7(a^3 +b^3 +c^3 )(a^4 +b^4 +c^4 ) ⇒ 10S_7 =7S_2 S_5 ⇒ 10(a^7 +b^7 +c^7 )=7(a^2 +b^2 +c^2 )(a^5 +b^5 +c^5 ) .........
$$\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{a}^{\mathrm{k}} +\mathrm{b}^{\mathrm{k}} +\mathrm{c}^{\mathrm{k}} =\mathrm{S}_{\mathrm{k}} \:\:;\:\:\mathrm{k}\in\mathbb{Z}^{+} \:\Rightarrow\:\mathrm{S}_{\mathrm{k}+\mathrm{3}} =\:\mathrm{abc}\:\mathrm{S}_{\mathrm{k}} +\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{S}_{\mathrm{2}} \mathrm{S}_{\mathrm{k}+\mathrm{1}} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{0}\:\:;\:\:\mathrm{a};\mathrm{b};\mathrm{c}\in\mathrm{I}\mathbb{R} \\ $$$$\Rightarrow\:\mathrm{S}_{\mathrm{0}} =\mathrm{3}\:\:\:\Rightarrow\:\mathrm{a}^{\mathrm{0}} +\mathrm{b}^{\mathrm{0}} +\mathrm{c}^{\mathrm{0}} =\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{S}_{\mathrm{1}} =\mathrm{0}\:\:\:\Rightarrow\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{S}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{2S}_{\mathrm{4}} \:\:\:\Rightarrow\:\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} +\mathrm{c}^{\mathrm{4}} \right) \\ $$$$\Rightarrow\:\mathrm{S}_{\mathrm{3}} =\mathrm{3abc}\:\Rightarrow\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} =\mathrm{3abc} \\ $$$$\Rightarrow\:\mathrm{6S}_{\mathrm{5}} =\mathrm{5S}_{\mathrm{2}} \mathrm{S}_{\mathrm{3}} \:\:\:\Rightarrow\:\mathrm{6}\left(\mathrm{a}^{\mathrm{5}} +\mathrm{b}^{\mathrm{5}} +\mathrm{c}^{\mathrm{5}} \right)=\mathrm{5}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)\left(\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\:\mathrm{6S}_{\mathrm{7}} =\mathrm{7S}_{\mathrm{3}} \mathrm{S}_{\mathrm{4}} \:\:\:\Rightarrow\:\mathrm{6}\left(\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} +\mathrm{c}^{\mathrm{7}} \right)=\mathrm{7}\left(\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \right)\left(\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} +\mathrm{c}^{\mathrm{4}} \right) \\ $$$$\Rightarrow\:\mathrm{10S}_{\mathrm{7}} =\mathrm{7S}_{\mathrm{2}} \mathrm{S}_{\mathrm{5}} \:\Rightarrow\:\mathrm{10}\left(\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} +\mathrm{c}^{\mathrm{7}} \right)=\mathrm{7}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \right)\left(\mathrm{a}^{\mathrm{5}} +\mathrm{b}^{\mathrm{5}} +\mathrm{c}^{\mathrm{5}} \right) \\ $$$$……… \\ $$
Commented by mathdanisur last updated on 03/Sep/21
Prove that if a+b+c=0 then: ((a^7 +b^7 +c^7 )/7) = ((a^5 +b^5 +c^5 )/5) = ((a^2 +b^2 +c^2 )/2)
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{0}\:\:\mathrm{then}: \\ $$$$\frac{\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} +\mathrm{c}^{\mathrm{7}} }{\mathrm{7}}\:=\:\frac{\mathrm{a}^{\mathrm{5}} +\mathrm{b}^{\mathrm{5}} +\mathrm{c}^{\mathrm{5}} }{\mathrm{5}}\:=\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by mr W last updated on 03/Sep/21
thanks alot sir!
$${thanks}\:{alot}\:{sir}! \\ $$
Commented by mr W last updated on 03/Sep/21
but ((a^7 +b^7 +c^7 )/7) = ((a^5 +b^5 +c^5 )/5) = ((a^2 +b^2 +c^2 )/2) is not true.
$${but}\: \\ $$$$\frac{\mathrm{a}^{\mathrm{7}} +\mathrm{b}^{\mathrm{7}} +\mathrm{c}^{\mathrm{7}} }{\mathrm{7}}\:=\:\frac{\mathrm{a}^{\mathrm{5}} +\mathrm{b}^{\mathrm{5}} +\mathrm{c}^{\mathrm{5}} }{\mathrm{5}}\:=\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${is}\:{not}\:{true}. \\ $$
Commented by Tawa11 last updated on 04/Sep/21
Nice
$$\mathrm{Nice} \\ $$
Answered by ajfour last updated on 03/Sep/21
a=2p, b=c=−p (((128−2))/(2(16+2)))=((126)/(36))=(7/2)
$${a}=\mathrm{2}{p},\:{b}={c}=−{p} \\ $$$$\frac{\left(\mathrm{128}−\mathrm{2}\right)}{\mathrm{2}\left(\mathrm{16}+\mathrm{2}\right)}=\frac{\mathrm{126}}{\mathrm{36}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 03/Sep/21
thanks sir!
$${thanks}\:{sir}! \\ $$
Answered by mr W last updated on 04/Sep/21
p_k =a^k +b^k +c^k e_1 =a+b+c e_2 =ab+bc+ca e_3 =abc p_1 =e_1 =0 p_2 =e_1 p_1 −2e_2 =−2e_2 p_3 =e_1 p_2 −e_2 p_1 +3e_3 =3e_3 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 =2e_2 ^2 p_5 =e_1 p_4 −e_2 p_3 +e_3 p_2 =−5e_2 e_3 p_6 =e_1 p_5 −e_2 p_4 +e_3 p_3 =−2e_2 ^3 +3e_3 ^2 p_7 =e_1 p_6 −e_2 p_5 +e_3 p_4 =7e_2 ^2 e_3 ⇒((a^7 +b^7 +c^7 )/(abc(a^4 +b^4 +c^4 )))=(p^7 /(e_3 p_4 ))=((7e_2 ^2 e_3 )/(e_3 ×2e_2 ^2 ))=(7/2) ((a^7 +b^7 +c^7 )/7)=e_2 ^2 e_3 =((−2e_2 )/2)×((−5e_2 e_3 )/5)=(p_2 /2)×(p_5 /5) ⇒((a^7 +b^7 +c^7 )/7)=((a^2 +b^2 +c^2 )/2)×((a^5 +b^5 +c^5 )/5)
$${p}_{{k}} ={a}^{{k}} +{b}^{{k}} +{c}^{{k}} \\ $$$${e}_{\mathrm{1}} ={a}+{b}+{c} \\ $$$${e}_{\mathrm{2}} ={ab}+{bc}+{ca} \\ $$$${e}_{\mathrm{3}} ={abc} \\ $$$$ \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{0} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} =−\mathrm{2}{e}_{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} =\mathrm{3}{e}_{\mathrm{3}} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} =−\mathrm{5}{e}_{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$${p}_{\mathrm{6}} ={e}_{\mathrm{1}} {p}_{\mathrm{5}} −{e}_{\mathrm{2}} {p}_{\mathrm{4}} +{e}_{\mathrm{3}} {p}_{\mathrm{3}} =−\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{3}} +\mathrm{3}{e}_{\mathrm{3}} ^{\mathrm{2}} \\ $$$${p}_{\mathrm{7}} ={e}_{\mathrm{1}} {p}_{\mathrm{6}} −{e}_{\mathrm{2}} {p}_{\mathrm{5}} +{e}_{\mathrm{3}} {p}_{\mathrm{4}} =\mathrm{7}{e}_{\mathrm{2}} ^{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} }{{abc}\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)}=\frac{{p}^{\mathrm{7}} }{{e}_{\mathrm{3}} {p}_{\mathrm{4}} }=\frac{\mathrm{7}{e}_{\mathrm{2}} ^{\mathrm{2}} {e}_{\mathrm{3}} }{{e}_{\mathrm{3}} ×\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\frac{{a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} }{\mathrm{7}}={e}_{\mathrm{2}} ^{\mathrm{2}} {e}_{\mathrm{3}} =\frac{−\mathrm{2}{e}_{\mathrm{2}} }{\mathrm{2}}×\frac{−\mathrm{5}{e}_{\mathrm{2}} {e}_{\mathrm{3}} }{\mathrm{5}}=\frac{{p}_{\mathrm{2}} }{\mathrm{2}}×\frac{{p}_{\mathrm{5}} }{\mathrm{5}} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} }{\mathrm{7}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}×\frac{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} }{\mathrm{5}} \\ $$
Commented by mathdanisur last updated on 04/Sep/21
thanks ser
$$\mathrm{thanks}\:\mathrm{ser} \\ $$

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