1.

If x4 + (1/x4) = 47 so find the value of x3 – (1/x3) 1. 5√52. 7√5 3. 8√54. 9√5

Answer» Correct Answer - Option 3 : 8√5

Given:

x4 + (1/x4) = 47

Formula used:

(x + y)2 = x2 + y2 + 2xy

(x – y)2 = x2 + y2 – 2xy

(x – y)3 =  x3 – y3 – 3xy (x – y)

Calculation:

x4 + (1/x4) = 47

⇒ [x2 + (1/x2)]2 = x4 + (1/x4) + 2 × x2 × (1/x2)

⇒  [x2 + (1/x2)]2 = 47 + 2

⇒ [x2 + (1/x2)]2 = 49

⇒ [x2 + (1/x2)] = 7

⇒ [x – (1/x)]2 = x2 + (1/x)2 – 2× x × (1/x)

⇒  [x – (1/x)]2 = 7 – 2

⇒ [x – (1/x)]2 = 5

⇒ x – (1/x) =√5

⇒ [x – (1/x)]3 =  x3 – (1/x)3 – 3 × x × (1/x) (x – 1/x)

⇒ x3 – (1/x)3 = [ x – (1/x)]3 + 3 (x – 1/x)

⇒ x3 – (1/x)3 = (√5) 3 + 3 × √5 

⇒ x3 – (1/x)3 = 5√5 + 3√5 

⇒ x3 – (1/x)3 = 8√5 

∴The value of x3 – (1/x)3 is 8√5 

Alternate method:

x4 + (1/x4) = 47 = P

x2 + (1/x2) = √(P + 2)

⇒ x2 + (1/x2) = √(47 + 2)

⇒ x2 + (1/x2) = 7 = Q

⇒ x – (1/x) = √(Q – 2)

⇒ x – (1/x) = √(7 – 2)

⇒ x – (1/x) = √5⇒ [x – (1/x)]3 =  x3 – (1/x)3 – 3 × x × (1/x) (x – 1/x)

⇒ x3 – (1/x)3 = [ x – (1/x)]3 + 3 (x – 1/x)

⇒ x3 – (1/x)3 = (√5) 3 + 3 × √5 

⇒ x3 – (1/x)3 = 5√5 + 3√5 

⇒ x3 – (1/x)3 = 8√5 

∴ The value of x3 – (1/x)3 is 8√5  


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