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If x2 – 7x + 1 = 0, then the value of \(\frac{{{x^6} + \;1}}{{{x^3}}}\;is\)
If x2 – 7x + 1 = 0, then the value of \(\frac{{{x^6} + \;1}}{{{x^3}}}\;is\)
1). 335
2). 322
3). 325
4). 328
1 answers
2 vote
Given,
x2 – 7x + 1 = 0
⇒ x2 + 1 = 7x
⇒ x + 1/x = 7 ….(1)
Squaring on both sides
⇒ (x + (1/x))2 = 49
⇒ x2 + 1/x2 = 49 – 2 (? (a + b)2 = a2 + 2ab + b2)
⇒ x2 + 1/x2 = 47 ….(2)
Now,
$(\frac{{{x^6} + \;1}}{{{x^3}}} = \;{x^3}\left( {\frac{{{x^3} + \frac{1}{{{x^3}}}}}{{{x^3}}}} \right) = {x^3} + \frac{1}{{{x^3}}} = \left( {x + \frac{1}{x}} \right)\left( {{x^2} - 1 + \frac{1}{{{x^2}}}} \right))$
= 7 × (47 – 1) = 7 × 46 = 322
∴ (x6 + 1)/x3 = 322