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Question Number 186105 by HeferH last updated on 01/Feb/23

$$a^2+1=a$$$$a^{12}+a^6+1=?$$

Answered by Rasheed.Sindhi last updated on 01/Feb/23

$$a^2+1=a;a^{12}+a^6+1=?$$$$$$$$a^2=a-1$$$$a^3=a^2-a=(a-1)-a=-1$$$$$$$$▸a^{12}+a^6+1={\left(a^3\right)}^4+{\left(a^3\right)}^2+1$$$$={(-1)}^4+{(-1)}^2+1=3$$

Answered by Rasheed.Sindhi last updated on 01/Feb/23

$$a^2-a+1=0\Rightarrow a+\frac{1}{a}=1;a\ne 0$$$${(a+\frac{1}{a})}^2=1^2\Rightarrow a^2+\frac{1}{a^2}=-1$$$${(a^2+\frac{1}{a^2})}^3={(-1)}^3$$$$a^6+\frac{1}{a^6}+3(a^2+\frac{1}{a^2})=-1$$$$a^6+\frac{1}{a^6}+3(-1)=-1\Rightarrow a^6+\frac{1}{a^6}=2$$$$a^{12}-2a^6+1=0$$$${(a^6-1)}^2=0$$$$a^6=1$$$$$$$$a^{12}+a^6+1=a^6(a^6+\frac{1}{a^6}+1)$$$$=\left(1\right)(2+1)=3$$

Answered by Rasheed.Sindhi last updated on 01/Feb/23

$$a^2+1=a;a^{12}+a^6+1=?$$$${(a^2+1)}^2=a^2$$$$a^4+2a^2+1=a^2$$$$a^4+1=-a^2$$$${(a^4+1)}^3={(-a^2)}^3$$$$a^{12}+1+3a^4(a^4+1)=-a^6$$$$a^{12}+1+3a^4(-a^2)=-a^6$$$$a^{12}-2a^6+1=0$$$${(a^6-1)}^2=0$$$$a^6=1$$$$▸a^{12}+a^6+1={\left(a^6\right)}^2+a^6+1$$$$={\left(1\right)}^2+1+1=3$$

Commented by HeferH last updated on 01/Feb/23

$${\left(thankyou\right)}^3!$$

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