• # question_answer The factors of ${{x}^{8}}+{{x}^{4}}+1$ are A)  $\left( {{x}^{4}}+1-{{x}^{2}} \right)\,\,\left( {{x}^{2}}+1+x \right)\,\,\left( {{x}^{2}}+1-x \right)$ B)  $\left( {{x}^{4}}+1-{{x}^{2}} \right)\,\,\left( {{x}^{2}}-1+x \right)\,\,\left( {{x}^{2}}+1+x \right)$ C)  $\left( {{x}^{4}}-1+{{x}^{2}} \right)\,\,\left( {{x}^{2}}-1+x \right)\,\,\left( {{x}^{2}}+1+x \right)$ D)  $\left( {{x}^{4}}-1+{{x}^{2}} \right)\,\,\left( {{x}^{2}}+1-x \right)\,\,\left( {{x}^{2}}+1+x \right)$

${{x}^{8}}+1+{{x}^{4}}={{({{x}^{4}})}^{2}}+{{(1)}^{2}}+{{x}^{4}}$ $={{({{x}^{4}}+1)}^{2}}-2{{x}^{2}}+{{x}^{4}}={{({{x}^{4}}+1)}^{2}}-{{({{x}^{2}})}^{2}}$ $=({{x}^{4}}+1+{{x}^{2}})({{x}^{4}}+1-{{x}^{2}})$ $=[{{({{x}^{2}})}^{2}}+{{(1)}^{2}}+{{x}^{2}}]\,\,[{{x}^{4}}-{{x}^{2}}+1]$ $=[{{({{x}^{2}}+1)}^{2}}-2{{x}^{2}}+{{x}^{2}}]\,\,[{{x}^{4}}-{{x}^{2}}+1]$ $=[{{({{x}^{2}}+1)}^{2}}-{{(x)}^{2}}]\,\,[{{x}^{4}}-{{x}^{2}}+1]$ $=({{x}^{2}}+1+x)({{x}^{2}}+1-x)({{x}^{4}}-{{x}^{2}}+1)$ $=({{x}^{2}}+x+1)({{x}^{2}}-x+1)({{x}^{4}}-{{x}^{2}}+1)$