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Question Number 212224 by RojaTaniya last updated on 07/Oct/24

$$a,b,c\in R$$$$a+b+c=1,a^2+b^2+c^2=1$$$$a^{10}+b^{10}+c^{10}=1,a^4+b^4+c^4=?$$

Answered by mr W last updated on 07/Oct/24

$$1^2=1+2(ab+bc+ca)$$$$\Rightarrow ab+bc+ca=0$$$$0^2=a^2{b}^2+b^2{c}^2+c^2{a}^2+2abc(a+b+c)$$$$\Rightarrow a^2{b}^2+b^2{c}^2+c^2{a}^2=-2abc=-2k$$$${(-2k)}^2=a^4{b}^4+b^4{c}^4+c^4{a}^4+2a^2{b}^2{c}^2(a^2+b^2+c^2)$$$$\Rightarrow a^4{b}^4+b^4{c}^4+c^4{a}^4=2k^2$$$$1^2=a^4+b^4+c^4+2(a^2{b}^2+b^2{c}^2+c^2{a}^2)$$$$\Rightarrow a^4+b^4+c^4=1+4k$$$$a^6+b^6+c^6+(a^2{b}^2+b^2{c}^2+c^2{a}^2)(a^2+b^2+c^2)-3a^2{b}^2{c}^2=(1+4k)\times 1$$$$\Rightarrow a^6+b^6+c^6=1+6k+3k^2$$$${(1+4k)}^2=a^8+b^8+c^8+2(a^4{b}^4+b^4{c}^4+c^4{a}^4)$$$$\Rightarrow a^8+b^8+c^8=1+8k+12k^2$$$$a^{10}+b^{10}+c^{10}+(a^2{b}^2+b^2{c}^2+c^2{a}^2)(a^6+b^6+c^6)-a^2{b}^2{c}^2(a^4+b^4+c^4)=(1+8k+12k^2)\times 1$$$$1+(-2k)(1+6k+3k^2)-k^2(1+4k)=(1+8k+12k^2)\times 1$$$$\Rightarrow k(2k+1)(k+2)=0$$$$\Rightarrow k=0,-\frac{1}{2},-2$$$$\Rightarrow a^4+b^4+c^4=1+4k=1,-1,-7$$$$ifa,b,c\in R:a^4+b^4+c^4=1$$

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