Question

If x² - 3x + 1 = 0, then find the value of (x⁶ + x⁴ + x² + 1)/x³ will be 
1. 18
2. 15
3. 21
4. 30

Answer 1

Correct Answer - Option 3 : 21

Given :- 

x2 - 3x + 1 = 0

Concept :- 

(a + b)³ = a³ + b³ + 3ab(a + b)

a³ + b³ = (a + b)³ - 3ab(a + b)

Calculation :- 

First divide given equation by x on both side

⇒ (x2 - 3x + 1)/x = (0/x)

⇒ (x²/x) - (3x/x) + (1/x) = 0

⇒ x - 3 + (1/x) = 0 

⇒ x + (1/x) = 3    ....(i)

Now, 

⇒ (x6 + x4 + x2 + 1)/x³

⇒ (x⁶/x³) + (x⁴/x³) + (x²/x³) + (1/x³)

⇒ x³ + x + (1/x) + (1/x³)

⇒ (x³ + (1/x³)) + (x + (1/x))    ....(ii)

Now cubing equation (i) on both side

⇒ (x + (1/x))³ = 3³

⇒ x3 + (1/x3) + 3 × x × (1/x) × (x + (1/x)) = 27

⇒ x3 + (1/x3) + 3 × 3 = 27  

⇒ x3 + (1/x3) = 27 -9 

⇒ x3 + (1/x3) = 18    ....(iii)

Now, 

Put the value of equation (iii) and (i) in equation (ii)

⇒ (x3 + (1/x3)) + (x + (1/x)) = 18 + 3 = 21

⇒ (x3 + (1/x3)) + (x + (1/x)) = 21

∴ (x3 + (1/x3)) + (x + (1/x)) is 21

Alternate method :- 

When 

x + (1/x) = a

⇒ x3 + (1/x3) = a³ - 3a

⇒ x + (1/x) = 3 

⇒ x3 + (1/x3) = 33 - 3 × 3

⇒ x3 + (1/x3) = 27 - 9

⇒ x3 + (1/x3) = 18

Now

⇒ x3 + x + (1/x) + (1/x3)

⇒ (x3 + (1/x3)) + (x + (1/x)) = 18 + 3= 21

⇒ (x3 + (1/x3)) + (x + (1/x)) = 21

∴ (x3 + (1/x3)) + (x + (1/x)) is 21