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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If $\frac{1}{a} (a^{2} + 1) = 3,$ $\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}$ then the value of $\frac{a^{6} + 1}{a^{3}}$ $\frac{{{a^6} + 1}}{{{a^3}}}$ = ?

19
218
327
41

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{1}{a} (a^{2} + 1) = 3 \\ \Rightarrow a + \frac{1}{a} = 3 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3. a . \frac{1}{a} (a + \frac{1}{a}) = 3^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 (3) = 3^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 27 - 9 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 18 \\ \Rightarrow \frac{a^{6} + 1}{a^{3}} = 18 \end{aligned}

$\eqalign{ & \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr & \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr}$

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then the value of

\frac{a^{6} + 1}{a^{3}}

$\frac{{{a^6} + 1}}{{{a^3}}}$ = ?", "text": "If

\frac{1}{a} (a^{2} + 1) = 3,

$\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}$ then the value of

\frac{a^{6} + 1}{a^{3}}

$\frac{{{a^6} + 1}}{{{a^3}}}$ = ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{1}{a} (a^{2} + 1) = 3 \\ \Rightarrow a + \frac{1}{a} = 3 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3. a . \frac{1}{a} (a + \frac{1}{a}) = 3^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 (3) = 3^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 27 - 9 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 18 \\ \Rightarrow \frac{a^{6} + 1}{a^{3}} = 18 \end{aligned}

$\eqalign{ & \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr & \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{1}{a} (a^{2} + 1) = 3 \\ \Rightarrow a + \frac{1}{a} = 3 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3. a . \frac{1}{a} (a + \frac{1}{a}) = 3^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 (3) = 3^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 27 - 9 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 18 \\ \Rightarrow \frac{a^{6} + 1}{a^{3}} = 18 \end{aligned}

$\eqalign{ & \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr & \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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