x3+1x3x3+1x3{x^3} + \frac{1}{{{x^3}}} is?" /> x3+1x3x3+1x3{x^3} + \frac{1}{{{x^3}}} is?" /> x3+1x3x3+1x3{x^3} + \frac{1}{{{x^3}}} is?" />

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If for a non - zero x, 3² + 5x + 3 = 0, then the value of $x^{3} + \frac{1}{x^{3}}$ ${x^3} + \frac{1}{{{x^3}}}$ is?

1 $\frac{10}{27}$ $\frac{{10}}{{27}}$
2 $- (\frac{10}{27})$ $- \left( {\frac{{10}}{{27}}} \right)$
3 $\frac{2}{3}$ $\frac{2}{3}$
4 $- (\frac{2}{3})$ $- \left( {\frac{2}{3}} \right)$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} 3 x^{2} + 5 x + 3 = 0 \\ 3 x^{2} + 3 = - 5 x \\ Divide by 3 x both sides \\ x + \frac{1}{x} = \frac{- 5}{3} \\ Then, x^{3} + \frac{1}{x^{3}} \\ = {(\frac{- 5}{3})}^{3} - 3 \times (\frac{- 5}{3}) \\ = \frac{- 125}{27} + 5 \\ = \frac{10}{27} \end{aligned}

$\eqalign{ & 3{x^2} + 5x + 3 = 0 \cr & 3{x^2} + 3 = - 5x \cr & {\text{Divide by }}3x{\text{ both sides}} \cr & x + \frac{1}{x} = \frac{{ - 5}}{3} \cr & {\text{Then, }}{x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\frac{{ - 5}}{3}} \right)^3} - 3 \times \left( {\frac{{ - 5}}{3}} \right) \cr & = \frac{{ - 125}}{{27}} + 5 \cr & = \frac{{10}}{{27}} \cr}$

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is?", "text": "If for a non - zero x, 3² + 5x + 3 = 0, then the value of

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

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

Solution:

\begin{aligned} 3 x^{2} + 5 x + 3 = 0 \\ 3 x^{2} + 3 = - 5 x \\ Divide by 3 x both sides \\ x + \frac{1}{x} = \frac{- 5}{3} \\ Then, x^{3} + \frac{1}{x^{3}} \\ = {(\frac{- 5}{3})}^{3} - 3 \times (\frac{- 5}{3}) \\ = \frac{- 125}{27} + 5 \\ = \frac{10}{27} \end{aligned}

$\eqalign{ & 3{x^2} + 5x + 3 = 0 \cr & 3{x^2} + 3 = - 5x \cr & {\text{Divide by }}3x{\text{ both sides}} \cr & x + \frac{1}{x} = \frac{{ - 5}}{3} \cr & {\text{Then, }}{x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\frac{{ - 5}}{3}} \right)^3} - 3 \times \left( {\frac{{ - 5}}{3}} \right) \cr & = \frac{{ - 125}}{{27}} + 5 \cr & = \frac{{10}}{{27}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 3 x^{2} + 5 x + 3 = 0 \\ 3 x^{2} + 3 = - 5 x \\ Divide by 3 x both sides \\ x + \frac{1}{x} = \frac{- 5}{3} \\ Then, x^{3} + \frac{1}{x^{3}} \\ = {(\frac{- 5}{3})}^{3} - 3 \times (\frac{- 5}{3}) \\ = \frac{- 125}{27} + 5 \\ = \frac{10}{27} \end{aligned}

$\eqalign{ & 3{x^2} + 5x + 3 = 0 \cr & 3{x^2} + 3 = - 5x \cr & {\text{Divide by }}3x{\text{ both sides}} \cr & x + \frac{1}{x} = \frac{{ - 5}}{3} \cr & {\text{Then, }}{x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\frac{{ - 5}}{3}} \right)^3} - 3 \times \left( {\frac{{ - 5}}{3}} \right) \cr & = \frac{{ - 125}}{{27}} + 5 \cr & = \frac{{10}}{{27}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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