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

$$f\left(x\right)=\frac{a^x+b^x+c^x}{x}witha+b+c=0$$$$provef\left(7\right)=f\left(5\right)\times f\left(2\right)$$

Answered by greougoury555 last updated on 05/Jan/23

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

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

$$whynotcontinue?$$$$whenyoumultiplef\left(2\right)withf\left(5\right),$$$$youshouldgetthesameasf\left(7\right).$$

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

$$f\left(1\right)=0$$$$f\left(2\right)=\frac{a^2+b^2+c^2}{2}=\frac{{(a+b+c)}^2-2(ab+bc+ca)}{2}$$$$f\left(2\right)=-ab-bc-ca$$$$f\left(3\right)=\frac{a^3+b^3+c^3}{3}$$$$f\left(3\right)-abc=\frac{a^3+b^3+c^3}{3}-abc$$$$=\frac{a^3+b^3+c^3-3abc}{3}$$$$=\frac{(a+b+c)(...)}{3}=0$$$$f\left(3\right)=abc$$$$Considerapolynomialequation$$$$withthreeroots:a,b,c$$$$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$$$$x^3-(-ab-bc-ca)x-abc=0$$$$x^3-xf\left(2\right)-f\left(3\right)=0$$$$....$$

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

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

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

$$fine!$$$$thanksfortheinterestsir!$$

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

$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{I}$$$${(a+b+c)}^2=a^2+b^2+c^2+2(ab+bc+ca)=0$$$$\Rightarrow a^2+b^2+c^2=-2(ab+bc+ca)$$$$\Rightarrow f\left(2\right)=\frac{a^2+b^2+c^2}{2}=-(ab+bc+ca)$$$${(a+b+c)}^3=a^3+b^3+c^2+3(a+b+c)(ab+bc+ca)-3abc=0$$$$\Rightarrow a^3+b^3+c^3=3abc$$$$\Rightarrow f\left(3\right)=\frac{a^3+b^3+c^3}{3}=abc$$$$(a^2+b^2+c^2)(a^3+b^3+c^2)=-6(ab+bc+ca)abc$$$$a^5+b^5+c^5+(a^2{b}^2+b^2{c}^2+c^2{a}^2)(a+b+c)-abc(ab+bc+ca)=-6(ab+bc+ca)abc$$$$\Rightarrow a^5+b^5+c^5=-5(ab+bc+ca)abc$$$$\Rightarrow f\left(5\right)=\frac{a^5+b^5+c^5}{5}=-(ab+bc+ca)abc$$$$(a^2+b^2+c^2)(a^5+b^5+c^5)=10{(ab+bc+ca)}^{2}abc$$$$a^7+b^7+c^7+(a^2{b}^2+b^2{c}^2+c^2{a}^2)(a^3+b^3+c^3)-a^2{b}^2{c}^2(a+b+c)=10{(ab+bc+ca)}^{2}abc$$$$a^7+b^7+c^7+3abc(a^2{b}^2+b^2{c}^2+c^2{a}^2)=10{(ab+bc+ca)}^{2}abc$$$$a^7+b^7+c^7+3abc[{(ab+bc+ca)}^2-2abc(a+b+c)]=10{(ab+bc+ca)}^{2}abc$$$$a^7+b^7+c^7+3abc{(ab+bc+ca)}^2=10{(ab+bc+ca)}^{2}abc$$$$\Rightarrow a^7+b^7+c^7=7{(ab+bc+ca)}^{2}abc$$$$\Rightarrow f\left(7\right)=\frac{a^7+b^7+c^7}{7}={(ab+bc+ca)}^{2}abc$$$$f\left(2\right)\times f\left(5\right)={(ab+bc+ca)}^{2}abc=f\left(7\right)✓$$

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

$$weseealsof\left(5\right)=f\left(3\right)\times f\left(2\right)$$

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

$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{I}\mathit{I}$$$$using{Newton}^{\prime }sIdentities$$$$let{p}_n=a^n+b^n+c^n$$$$e_1=a+b+c$$$$e_2=ab+bc+ca$$$$e_3=abc$$$$p_1=e_1=0$$$$p_2=e_1{p}_1-2e_2=-2e_2$$$$p_3=e_1{p}_2-e_2{p}_1+3e_3=3e_3$$$$p_4=e_1{p}_3-e_2{p}_2+e_3{p}_1=-e_2{p}_2=2e_2^2$$$$p_5=e_1{p}_4-e_2{p}_3+e_3{p}_2=-e_2{p}_3+e_3{p}_2=-5e_2{e}_3$$$$p_7=e_1{p}_6-e_2{p}_5+e_3{p}_4=-e_2{p}_5+e_3{p}_4=7e_2^2{e}_3$$$$f\left(n\right)=\frac{a^n+b^n+c^n}{n}=\frac{p_n}{n}$$$$f\left(7\right)=\frac{p_7}{7}=e_2^2{e}_3$$$$f\left(5\right)=\frac{p_5}{5}=-e_2{e}_3$$$$f\left(2\right)=\frac{p_2}{2}=-e_2$$$$f\left(5\right)\times f\left(2\right)=(-e_2{e}_3)(-e_2)=e_2^2{e}_3=f\left(7\right)$$

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