**If x – 1/x = √5, find the values of**

(i) x^{2} + 1/x^{2}

(ii) x + 1/x

(iii) x^{3} + 1/x^{3}

**Answer :**

**(i) x**^{2} + 1/x^{2} = (x – 1/x)^{2} + 2

^{2}+ 1/x

^{2}= (x – 1/x)

^{2}+ 2

Substituting the values

= (√5)^{2} + 2

= 5 + 2

= 7

**(ii) (x + 1/x)**^{2} = x^{2} + 1/x^{2} + 2

^{2}= x

^{2}+ 1/x

^{2}+ 2

Substituting the values

= 7 + 2

= 9

Here

(x + 1/x)^{2} = 9

So we get

(x + 1/x) = ± √9 = ± 3

**(iii) x**^{3} + 1/x^{3} = (x + 1/x)^{3} – 3x (1/x) (x + 1/x)

^{3}+ 1/x

^{3}= (x + 1/x)

^{3}– 3x (1/x) (x + 1/x)

Substituting the values

= (± 3)^{3} – 3 (± 3)

By further calculation

= (± 27) – (± 9)

= ± 18

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