1a(a2+1)=3,1a(a2+1)=3,\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}} then the value of a6+1a3" role="presentation">a6+1a3a6+1a3\frac{{{a^6} + 1}}{{{a^3}}} = ?" /> 1a(a2+1)=3,1a(a2+1)=3,\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}} then the value of a6+1a3" role="presentation">a6+1a3a6+1a3\frac{{{a^6} + 1}}{{{a^3}}} = ?" /> 1a(a2+1)=3,1a(a2+1)=3,\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}} then the value of a6+1a3" role="presentation">a6+1a3a6+1a3\frac{{{a^6} + 1}}{{{a^3}}} = ?" />
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