Question

If x⁴ + (1/x⁴) = 47 so find the value of x³ – (1/x³) 

1. 5√5
2. 7√5 
3. 8√5
4. 9√5

Answer 1

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