(a+1a)2=3,(a+1a)2=3,{\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}} the value of a3+1a3" role="presentation">a3+1a3a3+1a3{a^3} + \frac{1}{{{a^3}}} = ?" /> (a+1a)2=3,(a+1a)2=3,{\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}} the value of a3+1a3" role="presentation">a3+1a3a3+1a3{a^3} + \frac{1}{{{a^3}}} = ?" /> (a+1a)2=3,(a+1a)2=3,{\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}} the value of a3+1a3" role="presentation">a3+1a3a3+1a3{a^3} + \frac{1}{{{a^3}}} = ?" />

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

10
2 $3 (a + \frac{1}{a})$ $3\left( {a + \frac{1}{a}} \right)$
3 $3 (a^{2} + \frac{1}{a^{2}})$ $3\left( {{a^2} + \frac{1}{{{a^2}}}} \right)$
41

Answer:- 1

Explanation:-

Solution:

\begin{aligned} {(a + \frac{1}{a})}^{2} = 3 \\ \Rightarrow a + \frac{1}{a} = \sqrt{3} \\ Taking cube on both sides \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3. a . \frac{1}{a} (a + \frac{1}{a}) = 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 (\sqrt{3}) = 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 3 \sqrt{3} - 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 0 \end{aligned}

$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr}$

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the value of

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

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

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

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

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

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

Solution:

\begin{aligned} {(a + \frac{1}{a})}^{2} = 3 \\ \Rightarrow a + \frac{1}{a} = \sqrt{3} \\ Taking cube on both sides \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3. a . \frac{1}{a} (a + \frac{1}{a}) = 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 (\sqrt{3}) = 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 3 \sqrt{3} - 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 0 \end{aligned}

$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} {(a + \frac{1}{a})}^{2} = 3 \\ \Rightarrow a + \frac{1}{a} = \sqrt{3} \\ Taking cube on both sides \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3. a . \frac{1}{a} (a + \frac{1}{a}) = 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 (\sqrt{3}) = 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 3 \sqrt{3} - 3 \sqrt{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 0 \end{aligned}

$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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