Question

If x⁴ + 1/x⁴ =47 find the value of x³ + 1/x³



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Answer 1

Answer:-

Given:-

x⁴ + 1/x⁴ = 47

This can be written as:

⟹ (x²)² + (1/x²)² = 47

We know that,

a² + b² = (a + b)² - 2ab

So,

⟹ (x² + 1/x²)² - 2 * (x²) * (1/x²) = 47

⟹ (x² + 1/x²)² - 2 = 47

⟹ (x² + 1/x²)² = 47 + 2

⟹ (x² + 1/x²)² = 49

⟹ x² + 1/x² = √49

⟹ x² + 1/x² = 7

Again using a² + b² = (a + b)² - 2ab we get,

⟹ (x + 1/x)² - 2 (x)(1/x) = 7

⟹ (x + 1/x)² - 2 = 7

⟹ (x + 1/x)² = 7 + 2

⟹ (x + 1/x)² = 9

⟹ (x + 1/x) = √9

⟹ (x + 1/x) = 3 -- equation (1).

Now,

Cubing both sides we get,

⟹ (x + 1/x)³ = (3)³

using (a + b)³ = a³ + b³ + 3ab(a + b) we get,

⟹ x³ + 1/x³ + 3 (x) (1/x) (x + 1/x) = 27

Substituting x + 1/x = ± 3 [ from equation (1) ] in LHS we get,

⟹ x³ + 1/x³ + 3(3) = 27

⟹ x³ + 1/x³ + 9 = 27

⟹ x³ + 1/x³ = 27 - 9

⟹ x³ + 1/x³ = 18

∴ The value of x³ + 1/x³ is 18.

Answer 2

x⁴ + (1/x⁴) = 47

Adding and subtracting 2 on LHS,

x⁴ + (1/x⁴) + 2 - 2 = 47

[(x²)² + (1/x²)² + 2 × (1/x²) x²] = 47+2

( x² + 1/x² )² = 49

( x² + 1/x² )² = 7²

x² + 1/x² = 7 [neglecting the negative sign]

x² + 1/x² + 2 - 2 = 7

x² + 2.x².1/x² + 1/x² = 7+2

(x + 1/x)² = 9

(x+ 1/x)² = 3²

x + 1/x = 3 [neglecting the negative sign]

x³ + 1/x³ = (x+ 1/x)³ - 3.x.1/x(x + 1/x)

= (3)³ - 3(3)

= 27-9

= 18