Question
If $$\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}$$ then the value of $$\frac{{{a^6} + 1}}{{{a^3}}}$$ = ?
Answer: Option B
$$\eqalign{
& \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr
& \Rightarrow a + \frac{1}{a} = 3 \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr
& \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr} $$
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$$\eqalign{
& \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr
& \Rightarrow a + \frac{1}{a} = 3 \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr
& \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr} $$
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