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Question Number 212181 by RojaTaniya last updated on 05/Oct/24

Answered by A5T last updated on 05/Oct/24

$$a^2+b^2+c^2={(a+b+c)}^2-2(ab+bc+ca)=5$$$$\Rightarrow ab+bc+ca=2$$$$a^3+b^3+c^3-3abc=(a+b+c)[{(a+b+c)}^2-3(ab+bc+ca)]$$$$\Rightarrow 27-3abc=3[9-3\left(2\right)]=9\Rightarrow abc=6$$$$\Rightarrow a,b,carerootsof{x}^3-3x^2+2x-6=0$$$$\Rightarrow x^2(x-3)+2(x-3)=(x-3)(x^2+2)=0$$$$(a,b,c)=(3,i\sqrt{2},-i\sqrt{2})uptopermutation$$$$\Rightarrow a^{100}+b^{100}+c^{100}=3^{100}+2^{50}+2^{50}=3^{100}+2^{51}$$

Commented by RojaTaniya last updated on 05/Oct/24

$$Sirthanks$$

Answered by mr W last updated on 10/Oct/24

$$p_1=e_1=3$$$$p_2=e_1{p}_1-2e_2=5\Rightarrow e_2=2$$$$p_3=e_1{p}_2-e_2{p}_1+3e_3=27\Rightarrow e_3=6$$$$p_n=e_1{p}_{n-1}-e_2{p}_{n-2}+e_3{p}_{n-3}$$$$r^3-3r^2+2r-6=0$$$$r_1=3,r_{2,3}=\pm \sqrt{2}i$$$$p_n=3^n+2^{\frac{n}{2}}[i^n+{(-i)}^n]=3^n+2^{\frac{n}{2}+1}\cos \frac{n\pi }{2}$$$$\Rightarrow p_{100}=3^{100}+2^{51}$$

Commented by TonyCWX08 last updated on 08/Oct/24

$$Girand-{Newton}^{\prime }sIdentity?$$

Commented by mr W last updated on 08/Oct/24

$$yes$$

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