c+1c=3–√,c+1c=3,c + \frac{1}{c} = \sqrt 3 {\text{,}} then the value of c3+1c3" role="presentation">c3+1c3c3+1c3{c^3} + \frac{1}{{{c^3}}} is equal to?" /> c+1c=3–√,c+1c=3,c + \frac{1}{c} = \sqrt 3 {\text{,}} then the value of c3+1c3" role="presentation">c3+1c3c3+1c3{c^3} + \frac{1}{{{c^3}}} is equal to?" /> c+1c=3–√,c+1c=3,c + \frac{1}{c} = \sqrt 3 {\text{,}} then the value of c3+1c3" role="presentation">c3+1c3c3+1c3{c^3} + \frac{1}{{{c^3}}} is equal to?" />

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If $c + \frac{1}{c} = \sqrt{3},$ $c + \frac{1}{c} = \sqrt 3 {\text{,}}$ then the value of $c^{3} + \frac{1}{c^{3}}$ ${c^3} + \frac{1}{{{c^3}}}$ is equal to?

10
2 $\frac{3}{\sqrt{3}}$ $\frac{3}{{\sqrt 3 }}$
3 $\frac{1}{\sqrt{3}}$ $\frac{1}{{\sqrt 3 }}$
4 $\frac{6}{\sqrt{3}}$ $\frac{6}{{\sqrt 3 }}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} c + \frac{1}{c} = \sqrt{3} \\ On cubing both side \\ \Rightarrow {(c + \frac{1}{c})}^{3} = 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} + 3. c . \frac{1}{c} (c + \frac{1}{c}) = 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} + 3 \sqrt{3} = 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} = 3 \sqrt{3} - 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} = 0 \end{aligned}

$\eqalign{ & c + \frac{1}{c} = \sqrt 3 \cr & {\text{On cubing both side}} \cr & \Rightarrow {\left( {c + \frac{1}{c}} \right)^3} = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3.c.\frac{1}{c}\left( {c + \frac{1}{c}} \right) = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 0 \cr}$

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then the value of

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

${c^3} + \frac{1}{{{c^3}}}$ is equal to?", "text": "If

c + \frac{1}{c} = \sqrt{3},

$c + \frac{1}{c} = \sqrt 3 {\text{,}}$ then the value of

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

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

Solution:

\begin{aligned} c + \frac{1}{c} = \sqrt{3} \\ On cubing both side \\ \Rightarrow {(c + \frac{1}{c})}^{3} = 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} + 3. c . \frac{1}{c} (c + \frac{1}{c}) = 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} + 3 \sqrt{3} = 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} = 3 \sqrt{3} - 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} = 0 \end{aligned}

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

Solution:

\begin{aligned} c + \frac{1}{c} = \sqrt{3} \\ On cubing both side \\ \Rightarrow {(c + \frac{1}{c})}^{3} = 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} + 3. c . \frac{1}{c} (c + \frac{1}{c}) = 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} + 3 \sqrt{3} = 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} = 3 \sqrt{3} - 3 \sqrt{3} \\ \Rightarrow c^{3} + \frac{1}{c^{3}} = 0 \end{aligned}

$\eqalign{ & c + \frac{1}{c} = \sqrt 3 \cr & {\text{On cubing both side}} \cr & \Rightarrow {\left( {c + \frac{1}{c}} \right)^3} = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3.c.\frac{1}{c}\left( {c + \frac{1}{c}} \right) = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 0 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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