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Question Number 11477 by @ANTARES_VY last updated on 26/Mar/17

$$\frac{{\mathbf{a}}^8+{\mathbf{a}}^4+1}{{\mathbf{a}}^6+1}.$$$$\mathbf{solves}....$$

Answered by sm3l2996 last updated on 26/Mar/17

$${\mathrm{a}}^8+{\mathrm{a}}^4+1={\mathrm{a}}^2({\mathrm{a}}^6+1)+{\mathrm{a}}^4-{\mathrm{a}}^2+1$$$$\frac{{\mathrm{a}}^8+{\mathrm{a}}^4+1}{{\mathrm{a}}^6+1}={\mathrm{a}}^2+\frac{{\mathrm{a}}^4-{\mathrm{a}}^2+1}{{\mathrm{a}}^6+1}$$$$\frac{{\mathrm{a}}^4-{\mathrm{a}}^2+1}{{\mathrm{a}}^6+1}=\frac{{\mathrm{a}}^4-{\mathrm{a}}^2+1}{({\mathrm{a}}^2+1)({\mathrm{a}}^4-{\mathrm{a}}^2+1)}=\frac{1}{{\mathrm{a}}^2+1}$$$$\frac{{\mathrm{a}}^8+{\mathrm{a}}^4+1}{{\mathrm{a}}^6+1}={\mathrm{a}}^2+\frac{1}{{\mathrm{a}}^2+1}$$

Commented by @ANTARES_VY last updated on 27/Mar/17

$$\mathbf{T}\mathrm{henks}.$$

Commented by @ANTARES_VY last updated on 27/Mar/17

$$\mathbf{T}\mathrm{henks}.$$

Commented by @ANTARES_VY last updated on 27/Mar/17

$$\mathbf{T}\mathrm{henks}.$$

Commented by @ANTARES_VY last updated on 27/Mar/17

$$\mathbf{T}\mathrm{henks}.$$

Commented by sma3l2996 last updated on 27/Mar/17

$$youwelcome$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/Mar/17

$$a^2=p\Rightarrow \frac{p^4+p^2+1}{p^3+1}=\frac{\frac{p^6-1}{p^2-1}}{p^3+1}=\frac{(p^3+1)(p^3-1)}{(p^3+1)(p^2-1)}$$$$=\frac{(p-1)(p^2+p+1)}{(p-1)(p+1)}=\frac{p^2+p+1}{p+1}=\frac{a^4+a^2+1}{a^2+1}$$$$\frac{{(a^2+1)}^2-a^2}{a^2+1}=\frac{(a^2+a+1)(a^2-a+1)}{a^2+1}$$

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