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Question Number 184351 by mr W last updated on 05/Jan/23

f(x)=((a^x +b^x +c^x )/x) with a+b+c=0 prove f(7)=f(5)×f(2)

$${f}\left({x}\right)=\frac{{a}^{{x}} +{b}^{{x}} +{c}^{{x}} }{{x}}\:{with}\:{a}+{b}+{c}=\mathrm{0} \\ $$$${prove}\:{f}\left(\mathrm{7}\right)={f}\left(\mathrm{5}\right)×{f}\left(\mathrm{2}\right) \\ $$

Answered by greougoury555 last updated on 05/Jan/23

f(x)=((a^x +b^x +(−a−b)^x )/x) { ((f(7)=((a^7 +b^7 −(a+b)^7 )/7))),((f(5)=((a^5 +b^5 −(a+b)^5 )/5))),((f(2)=((a^2 +b^2 +(a+b)^2 )/2))) :} f(2)=((2(a^2 +b^2 +ab))/2) = a^2 +b^2 +ab f(5)=((−(5a^4 b+10a^3 b^2 +10a^2 b^3 +5ab^4 ))/5) f(5)=−ab(a^3 +2a^2 b+2ab^2 +b^3 ) f(7)=((−(7a^6 b+21a^5 b^2 +35a^4 b^3 +35a^3 b^4 +21a^2 b^5 +7ab^6 ))/7) f(7)=−ab(a^5 +3a^4 b+5a^3 b^2 +5a^2 b^3 +3ab^4 +b^5 )

$$\:{f}\left({x}\right)=\frac{{a}^{{x}} +{b}^{{x}} +\left(−{a}−{b}\right)^{{x}} }{{x}} \\ $$$$\:\begin{cases}{{f}\left(\mathrm{7}\right)=\frac{{a}^{\mathrm{7}} +{b}^{\mathrm{7}} −\left({a}+{b}\right)^{\mathrm{7}} }{\mathrm{7}}}\\{{f}\left(\mathrm{5}\right)=\frac{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} −\left({a}+{b}\right)^{\mathrm{5}} }{\mathrm{5}}}\\{{f}\left(\mathrm{2}\right)=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{2}}}\end{cases} \\ $$$$\:{f}\left(\mathrm{2}\right)=\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}\right)}{\mathrm{2}}\:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab} \\ $$$$\:{f}\left(\mathrm{5}\right)=\frac{−\left(\mathrm{5}{a}^{\mathrm{4}} {b}+\mathrm{10}{a}^{\mathrm{3}} {b}^{\mathrm{2}} +\mathrm{10}{a}^{\mathrm{2}} {b}^{\mathrm{3}} +\mathrm{5}{ab}^{\mathrm{4}} \right)}{\mathrm{5}} \\ $$$$\:{f}\left(\mathrm{5}\right)=−{ab}\left({a}^{\mathrm{3}} +\mathrm{2}{a}^{\mathrm{2}} {b}+\mathrm{2}{ab}^{\mathrm{2}} +{b}^{\mathrm{3}} \right) \\ $$$$\:{f}\left(\mathrm{7}\right)=\frac{−\left(\mathrm{7}{a}^{\mathrm{6}} {b}+\mathrm{21}{a}^{\mathrm{5}} {b}^{\mathrm{2}} +\mathrm{35}{a}^{\mathrm{4}} {b}^{\mathrm{3}} +\mathrm{35}{a}^{\mathrm{3}} {b}^{\mathrm{4}} +\mathrm{21}{a}^{\mathrm{2}} {b}^{\mathrm{5}} +\mathrm{7}{ab}^{\mathrm{6}} \right)}{\mathrm{7}} \\ $$$$\:{f}\left(\mathrm{7}\right)=−{ab}\left({a}^{\mathrm{5}} +\mathrm{3}{a}^{\mathrm{4}} {b}+\mathrm{5}{a}^{\mathrm{3}} {b}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} {b}^{\mathrm{3}} +\mathrm{3}{ab}^{\mathrm{4}} +{b}^{\mathrm{5}} \right) \\ $$

Commented by mr W last updated on 06/Jan/23

why not continue? when you multiple f(2) with f(5), you should get the same as f(7).

$${why}\:{not}\:{continue}? \\ $$$${when}\:{you}\:{multiple}\:{f}\left(\mathrm{2}\right)\:{with}\:{f}\left(\mathrm{5}\right), \\ $$$${you}\:{should}\:{get}\:{the}\:{same}\:{as}\:{f}\left(\mathrm{7}\right). \\ $$

Answered by Rasheed.Sindhi last updated on 05/Jan/23

f(1)=0 f(2)=((a^2 +b^2 +c^2 )/2)=(((a+b+c)^2 −2(ab+bc+ca))/2) f(2)=−ab−bc−ca f(3)=((a^3 +b^3 +c^3 )/3) f(3)−abc=((a^3 +b^3 +c^3 )/3)−abc =((a^3 +b^3 +c^3 −3abc)/3) =(((a+b+c)(...))/3)=0 f(3)=abc Consider a polynomial equation with three roots: a,b,c x^3 −(a+b+c)x^2 +(ab+bc+ca)x−abc=0 x^3 −(−ab−bc−ca)x−abc=0 x^3 −xf(2)−f(3)=0 ....

$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right)}{\mathrm{2}} \\ $$$${f}\left(\mathrm{2}\right)=−{ab}−{bc}−{ca} \\ $$$${f}\left(\mathrm{3}\right)=\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${f}\left(\mathrm{3}\right)−{abc}=\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{\mathrm{3}}−{abc} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:=\frac{\left({a}+{b}+{c}\right)\left(...\right)}{\mathrm{3}}=\mathrm{0} \\ $$$${f}\left(\mathrm{3}\right)={abc} \\ $$$${Consider}\:{a}\:{polynomial}\:{equation} \\ $$$${with}\:{three}\:{roots}:\:{a},{b},{c} \\ $$$${x}^{\mathrm{3}} −\left({a}+{b}+{c}\right){x}^{\mathrm{2}} +\left({ab}+{bc}+{ca}\right){x}−{abc}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\left(−{ab}−{bc}−{ca}\right){x}−{abc}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −{xf}\left(\mathrm{2}\right)−{f}\left(\mathrm{3}\right)=\mathrm{0} \\ $$$$.... \\ $$

Answered by Rasheed.Sindhi last updated on 06/Jan/23

•a+b+c=0⇒c=−a−b ▶f(2)=((a^2 +b^2 +c^2 )/2)=(((a+b+c)^2 −2(ab+bc+ca))/2) =−ab−bc−ca=−ab−c(a+b) =−ab−(a+b)(−a−b) =−ab+(a+b)^2 •f(2) =a^2 +ab+b^2 ▶f(5)=((a^5 +b^5 +c^5 )/5)=((a^5 +b^5 +(−a−b)^5 )/5) =((a^5 +b^5 −(a+b)^5 )/5) =((−(5a^4 b+10a^3 b^2 +10a^2 b^3 +5ab^4 ))/5) •f(5) =−(a^4 b+2a^3 b^2 +2a^2 b^3 +ab^4 ) ▶ f(7)=((a^7 +b^7 +c^7 )/7)=((a^7 +b^7 +(−a−b)^7 )/7) =((a^7 +b^7 −(a+b)^7 )/7) •f(7)=−(a^6 b+3a^5 b^2 +5a^4 b^3 +5a^3 b^4 +3a^2 b^5 +ab^6 ) ▶ f(2)f(5): −a^4 b−2a^3 b^2 −2a^2 b^3 −ab^4 × (a^2 +ab+b^2 ) _(−) −a^6 b−2a^5 b^2 −2a^4 b^3 −a^3 b^4 −a^5 b^2 −2a^4 b^3 −2a^3 b^4 −a^2 b^5 − a^4 b^3 −2a^3 b^4 −2a^2 b^5 −ab^6 _(−) −a^6 b−3a^5 b^2 −5a^4 b^3 −5a^3 b^4 −3a^2 b^5 −ab^6 =f(7)

$$ \\ $$$$\bullet{a}+{b}+{c}=\mathrm{0}\Rightarrow{c}=−{a}−{b} \\ $$$$\blacktriangleright{f}\left(\mathrm{2}\right)=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−{ab}−{bc}−{ca}=−{ab}−{c}\left({a}+{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−{ab}−\left({a}+{b}\right)\left(−{a}−{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−{ab}+\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\bullet{f}\left(\mathrm{2}\right)\:={a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \\ $$$$\blacktriangleright{f}\left(\mathrm{5}\right)=\frac{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} }{\mathrm{5}}=\frac{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +\left(−{a}−{b}\right)^{\mathrm{5}} }{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} −\left({a}+{b}\right)^{\mathrm{5}} }{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{−\left(\mathrm{5}{a}^{\mathrm{4}} {b}+\mathrm{10}{a}^{\mathrm{3}} {b}^{\mathrm{2}} +\mathrm{10}{a}^{\mathrm{2}} {b}^{\mathrm{3}} +\mathrm{5}{ab}^{\mathrm{4}} \right)}{\mathrm{5}} \\ $$$$\bullet{f}\left(\mathrm{5}\right)\:=−\left({a}^{\mathrm{4}} {b}+\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{3}} +{ab}^{\mathrm{4}} \right) \\ $$$$\blacktriangleright\:{f}\left(\mathrm{7}\right)=\frac{{a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} }{\mathrm{7}}=\frac{{a}^{\mathrm{7}} +{b}^{\mathrm{7}} +\left(−{a}−{b}\right)^{\mathrm{7}} }{\mathrm{7}} \\ $$$$\:\:\:\:\:\:\:=\frac{{a}^{\mathrm{7}} +{b}^{\mathrm{7}} −\left({a}+{b}\right)^{\mathrm{7}} }{\mathrm{7}} \\ $$$$\bullet{f}\left(\mathrm{7}\right)=−\left({a}^{\mathrm{6}} {b}+\mathrm{3}{a}^{\mathrm{5}} {b}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{4}} {b}^{\mathrm{3}} +\mathrm{5}{a}^{\mathrm{3}} {b}^{\mathrm{4}} +\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{5}} +{ab}^{\mathrm{6}} \right) \\ $$$$\blacktriangleright\:{f}\left(\mathrm{2}\right){f}\left(\mathrm{5}\right): \\ $$$$\:\:−{a}^{\mathrm{4}} {b}−\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{3}} −{ab}^{\mathrm{4}} \:\:\: \\ $$$$\underset{−} {\:\:\:\:\:\:\:×\:\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} \\ $$$$\:\:\:\:\:\:−{a}^{\mathrm{6}} {b}−\mathrm{2}{a}^{\mathrm{5}} {b}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{4}} {b}^{\mathrm{3}} −{a}^{\mathrm{3}} {b}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{a}^{\mathrm{5}} {b}^{\mathrm{2}} \:\:−\mathrm{2}{a}^{\mathrm{4}} {b}^{\mathrm{3}} −\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{5}} \\ $$$$\underset{−} {\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\:\:\:{a}^{\mathrm{4}} {b}^{\mathrm{3}} −\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{5}} −{ab}^{\mathrm{6}} \:\:\:} \\ $$$$−{a}^{\mathrm{6}} {b}−\mathrm{3}{a}^{\mathrm{5}} {b}^{\mathrm{2}} −\mathrm{5}{a}^{\mathrm{4}} {b}^{\mathrm{3}} −\mathrm{5}{a}^{\mathrm{3}} {b}^{\mathrm{4}} −\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{5}} −{ab}^{\mathrm{6}} \\ $$$$={f}\left(\mathrm{7}\right) \\ $$

Commented by mr W last updated on 06/Jan/23

fine! thanks for the interest sir!

$${fine}! \\ $$$${thanks}\:{for}\:{the}\:{interest}\:{sir}! \\ $$

Answered by mr W last updated on 06/Jan/23

Method I (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ca)=0 ⇒a^2 +b^2 +c^2 =−2(ab+bc+ca) ⇒f(2)=((a^2 +b^2 +c^2 )/2)=−(ab+bc+ca) (a+b+c)^3 =a^3 +b^3 +c^2 +3(a+b+c)(ab+bc+ca)−3abc=0 ⇒a^3 +b^3 +c^3 =3abc ⇒f(3)=((a^3 +b^3 +c^3 )/3)=abc (a^2 +b^2 +c^2 )(a^3 +b^3 +c^2 )=−6(ab+bc+ca)abc a^5 +b^5 +c^5 +(a^2 b^2 +b^2 c^2 +c^2 a^2 )(a+b+c)−abc(ab+bc+ca)=−6(ab+bc+ca)abc ⇒a^5 +b^5 +c^5 =−5(ab+bc+ca)abc ⇒f(5)=((a^5 +b^5 +c^5 )/5)=−(ab+bc+ca)abc (a^2 +b^2 +c^2 )(a^5 +b^5 +c^5 )=10(ab+bc+ca)^2 abc a^7 +b^7 +c^7 +(a^2 b^2 +b^2 c^2 +c^2 a^2 )(a^3 +b^3 +c^3 )−a^2 b^2 c^2 (a+b+c)=10(ab+bc+ca)^2 abc a^7 +b^7 +c^7 +3abc(a^2 b^2 +b^2 c^2 +c^2 a^2 )=10(ab+bc+ca)^2 abc a^7 +b^7 +c^7 +3abc[(ab+bc+ca)^2 −2abc(a+b+c)]=10(ab+bc+ca)^2 abc a^7 +b^7 +c^7 +3abc(ab+bc+ca)^2 =10(ab+bc+ca)^2 abc ⇒a^7 +b^7 +c^7 =7(ab+bc+ca)^2 abc ⇒f(7)=((a^7 +b^7 +c^7 )/7)=(ab+bc+ca)^2 abc f(2)×f(5)=(ab+bc+ca)^2 abc=f(7) ✓

$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =−\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{2}\right)=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}=−\left({ab}+{bc}+{ca}\right) \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{2}} +\mathrm{3}\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)−\mathrm{3}{abc}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc} \\ $$$$\Rightarrow{f}\left(\mathrm{3}\right)=\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{\mathrm{3}}={abc} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{2}} \right)=−\mathrm{6}\left({ab}+{bc}+{ca}\right){abc} \\ $$$${a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} +\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)\left({a}+{b}+{c}\right)−{abc}\left({ab}+{bc}+{ca}\right)=−\mathrm{6}\left({ab}+{bc}+{ca}\right){abc} \\ $$$$\Rightarrow{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} =−\mathrm{5}\left({ab}+{bc}+{ca}\right){abc} \\ $$$$\Rightarrow{f}\left(\mathrm{5}\right)=\frac{{a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} }{\mathrm{5}}=−\left({ab}+{bc}+{ca}\right){abc} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{5}} +{b}^{\mathrm{5}} +{c}^{\mathrm{5}} \right)=\mathrm{10}\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} {abc} \\ $$$${a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} +\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \left({a}+{b}+{c}\right)=\mathrm{10}\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} {abc} \\ $$$${a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} +\mathrm{3}{abc}\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \right)=\mathrm{10}\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} {abc} \\ $$$${a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} +\mathrm{3}{abc}\left[\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} −\mathrm{2}{abc}\left({a}+{b}+{c}\right)\right]=\mathrm{10}\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} {abc} \\ $$$${a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} +\mathrm{3}{abc}\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} =\mathrm{10}\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} {abc} \\ $$$$\Rightarrow{a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} =\mathrm{7}\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} {abc} \\ $$$$\Rightarrow{f}\left(\mathrm{7}\right)=\frac{{a}^{\mathrm{7}} +{b}^{\mathrm{7}} +{c}^{\mathrm{7}} }{\mathrm{7}}=\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} {abc} \\ $$$${f}\left(\mathrm{2}\right)×{f}\left(\mathrm{5}\right)=\left({ab}+{bc}+{ca}\right)^{\mathrm{2}} {abc}={f}\left(\mathrm{7}\right)\:\checkmark \\ $$

Commented by mr W last updated on 06/Jan/23

we see also f(5)=f(3)×f(2)

$${we}\:{see}\:{also}\:{f}\left(\mathrm{5}\right)={f}\left(\mathrm{3}\right)×{f}\left(\mathrm{2}\right) \\ $$

Answered by mr W last updated on 06/Jan/23

Method II using Newton′s Identities let p_n =a^n +b^n +c^n 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 =−e_2 p_2 =2e_2 ^2 p_5 =e_1 p_4 −e_2 p_3 +e_3 p_2 =−e_2 p_3 +e_3 p_2 =−5e_2 e_3 p_7 =e_1 p_6 −e_2 p_5 +e_3 p_4 =−e_2 p_5 +e_3 p_4 =7e_2 ^2 e_3 f(n)=((a^n +b^n +c^n )/n)=(p_n /n) f(7)=(p_7 /7)=e_2 ^2 e_3 f(5)=(p_5 /5)=−e_2 e_3 f(2)=(p_2 /2)=−e_2 f(5)×f(2)=(−e_2 e_3 )(−e_2 )=e_2 ^2 e_3 =f(7)

$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$${using}\:{Newton}'{s}\:{Identities} \\ $$$${let}\:{p}_{{n}} ={a}^{{n}} +{b}^{{n}} +{c}^{{n}} \\ $$$${e}_{\mathrm{1}} ={a}+{b}+{c} \\ $$$${e}_{\mathrm{2}} ={ab}+{bc}+{ca} \\ $$$${e}_{\mathrm{3}} ={abc} \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{0} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} =−\mathrm{2}{e}_{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} =\mathrm{3}{e}_{\mathrm{3}} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =−{e}_{\mathrm{2}} {p}_{\mathrm{2}} =\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} =−{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} =−\mathrm{5}{e}_{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$${p}_{\mathrm{7}} ={e}_{\mathrm{1}} {p}_{\mathrm{6}} −{e}_{\mathrm{2}} {p}_{\mathrm{5}} +{e}_{\mathrm{3}} {p}_{\mathrm{4}} =−{e}_{\mathrm{2}} {p}_{\mathrm{5}} +{e}_{\mathrm{3}} {p}_{\mathrm{4}} =\mathrm{7}{e}_{\mathrm{2}} ^{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$${f}\left({n}\right)=\frac{{a}^{{n}} +{b}^{{n}} +{c}^{{n}} }{{n}}=\frac{{p}_{{n}} }{{n}} \\ $$$${f}\left(\mathrm{7}\right)=\frac{{p}_{\mathrm{7}} }{\mathrm{7}}={e}_{\mathrm{2}} ^{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$${f}\left(\mathrm{5}\right)=\frac{{p}_{\mathrm{5}} }{\mathrm{5}}=−{e}_{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$${f}\left(\mathrm{2}\right)=\frac{{p}_{\mathrm{2}} }{\mathrm{2}}=−{e}_{\mathrm{2}} \\ $$$${f}\left(\mathrm{5}\right)×{f}\left(\mathrm{2}\right)=\left(−{e}_{\mathrm{2}} {e}_{\mathrm{3}} \right)\left(−{e}_{\mathrm{2}} \right)={e}_{\mathrm{2}} ^{\mathrm{2}} {e}_{\mathrm{3}} ={f}\left(\mathrm{7}\right) \\ $$

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