(a^4+a^2+1)...

Ikosha212 @Ikosha212

March 2022 1 20 Report

сократить дробь (a^2-a+1)/(a^4+a^2+1)

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NNNLLL54

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$\frac{a^2-a+1}{a^4+a^2+1}=\frac{a^2-a+1}{(a^2)^2+1\cdot a^2+1^2}=\frac{(a^2-1)\cdot (a^2-a+1)}{(a^2-1)((a^2)^2+1\cdot a^2+1^2)}=\\\\=\frac{(a-1)(a+1)(a^2-a+1)}{(a^2)^3-1^3}=\frac{(a-1)\cdot (a^3+1^3)}{a^6-1}=\frac{(a-1)(a^3+1)}{(a^3)^2-1^2}=\\\\=\frac{(a-1)(a^3+1)}{(a^3-1)(a^3+1)}=\frac{a-1}{a^3-1}=\frac{a-1}{(a-1)(a^2+a+1)}=\frac{1}{a^2+a+1}\\\\\\\star \; \; \; x^3-y^3=(x-y)(x^2+xy+y^2)\; \; \; \star \\\\\star \; \; \; x^3+y^3=(x+y)(x^2-xy+y^2)\; \; \; \star$

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