Question

If x²+1/x²=66, find the value of (I)x³-1/x³(ii)x⁶-1/x⁶​

Answer 1

Solution 1 :-

→ x² + 1/x² = 66

Subtracting 2 both sides ,

→ x² + 1/x² - 2 = 66 - 2

→ x² + 1/x² - 2 * x * 1/x = 64

comparing with a² + b² - 2ab = (a - b)² now,

→ (x - 1/x)² = 8²

Square root both sides,

→ (x - 1/x) = 8

Therefore,

→ (x³ - 1/x³)

using a³ - b³ = (a - b)(a² + b² + ab)

→ (x - 1/x)(x² + 1/x² + 1)

→ 8 * (66 + 1)

→ 8 * 67

→ 536 (Ans.)

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Solution 2 :-

→ x² + 1/x² = 66

Adding 2 both sides ,

→ x² + 1/x² - 2 = 66 + 2

→ x² + 1/x² + 2 * x * 1/x = 68

comparing with a² + b² + 2ab = (a + b)² now,

→ (x + 1/x)² = 68

Square root both sides,

→ (x + 1/x) = √68 = 2√17

Therefore,

→ (x³ + 1/x³)

using a³ + b³ = (a + b)(a² + b² - ab)

→ 2√17 * (66 - 1)

→ 2√17 * 65

→ 130√17

Hence,

→ (x⁶-1/x⁶)

→ (x³)² - (1/x³)²

using (a² - b²) = (a + b)(a - b)

→ (x³ + 1/x³)(x³ - 1/x³)

Putting both values now,

→ 130√17 * 536

→ 69,680√17 (Ans.)

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