Question

If t² - 4t + 1 = 0, then the value of \(t^3+\frac{1}{t^3}\)is
1. 44
2. 48
3. 52
4. 64

Answer 1

Correct Answer - Option 3 : 52

Given:

t2 – 4t + 1 = 0

Formula used:

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculation:

Given expression is t2 – 4t + 1 = 0

⇒ t2 + 1 = 4t

⇒ t + 1/t = 4      ----(1)

Now, Taking cube both side

(t + 1/t)³ = 4³

⇒ t³ + 1/t³ + 3(t × 1/t)(t + 1/t) = 64

⇒ t3 + 1/t3 + 3(t + 1/t) = 64

⇒ t3 + 1/t3 + 3 × 4 = 64      ----(From eq (1))

⇒ t3 + 1/t3 = 64 – 12 

∴ t3 + 1/t3 = 52