If, for a non-zero x, 5x2 + 7x + 5 = 0, then the value of \({x^3} + \frac{1}{{{x^3}}}\) is:

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SSC CGL 2022 Tier-I Official Paper (Held On : 06 Dec 2022 Shift 4)
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  1. \(\frac{{496}}{{125}}\)
  2. \(\frac{{532}}{{343}}\)
  3. \(\frac{{125}}{{532}}\)
  4. \(\frac{{182}}{{125}}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{{182}}{{125}}\)
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Detailed Solution

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Given:

5x2 + 7x + 5 = 0

Calculations:

According to question,

5x2 + 7x + 5 = 0

⇒5x² + 5 = - 7x

⇒5x(x + 1/x) = - 7x

⇒x + 1/x = -7/5

so, x³ + 1/x³

= (-7/5)³ - 3 × (-7/5)

= - 343/125 + 21/5

= (-343 + 525 )/125

= 182/125

Hence, The Required value is 182/125.

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