a+1a=3,a+1a=3,a + \frac{1}{a} = 3, then the value of a3+1a3" role="presentation">a3+1a3a3+1a3{a^3} + \frac{1}{{{a^3}}} is?" /> a+1a=3,a+1a=3,a + \frac{1}{a} = 3, then the value of a3+1a3" role="presentation">a3+1a3a3+1a3{a^3} + \frac{1}{{{a^3}}} is?" /> a+1a=3,a+1a=3,a + \frac{1}{a} = 3, then the value of a3+1a3" role="presentation">a3+1a3a3+1a3{a^3} + \frac{1}{{{a^3}}} is?" />

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If $a + \frac{1}{a} = 3,$ $a + \frac{1}{a} = 3,$ then the value of $a^{3} + \frac{1}{a^{3}}$ ${a^3} + \frac{1}{{{a^3}}}$ is?

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given , a + \frac{1}{a} = 3 \\ Cube both sides \\ a^{3} + \frac{1}{a^{3}} + 3 \times a \times \frac{1}{a} (a + \frac{1}{a}) = {(3)}^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 \times 3 = 27 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 27 - 9 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 18 \end{aligned}

$\eqalign{ & {\text{Given , }}a + \frac{1}{a} = 3 \cr & {\text{Cube both sides}} \cr & {a^3} + \frac{1}{{{a^3}}} + 3 \times a \times \frac{1}{a}\left( {a + \frac{1}{a}} \right) = {\left( 3 \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr}$

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then the value of

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

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

a + \frac{1}{a} = 3,

$a + \frac{1}{a} = 3,$ then the value of

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

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

Solution: