Question

$\rm If \ x+\dfrac{1}{x}=3, \ find \ x^8 + \dfrac{1}{x^8}$​

Answer 1

We've been given that:

$\rm \longmapsto \ x + \dfrac{1}{x} = 3$

Squaring on both sides we get:

$\rm \longmapsto \ \Bigg(x + \dfrac{1}{x} \Bigg)^2 = \big(3\big)^2$

Applying the identity (a + b)² = a² + b² + 2ab we get:

$\rm \longmapsto \ (x)^2 + \Bigg(\dfrac{1}{x} \Bigg)^2 + 2\big(x\big)\Bigg(\dfrac{1}{x} \Bigg) = 9$

$\rm \longmapsto \ x^2 + \dfrac{1}{x^2} + 2 = 9$

$\rm \longmapsto \ x^2 + \dfrac{1}{x^2} = 9 - 2$

$\rm \longmapsto \ x^2 + \dfrac{1}{x^2} = 7$

Square on both sides again.

$\rm \longmapsto \ \Bigg(x^2 + \dfrac{1}{x^2} \Bigg)^2 = \big(7\big)^2$

Applying the identity (a + b)² = a² + b² + 2ab we get:

$\rm \longmapsto \ \Big(x^2\Big)^2 + \Bigg(\dfrac{1}{x^2}\Bigg)^2 + 2\Big(x^2\Big)\Bigg(\dfrac{1}{x^2}\Bigg) = 49$

$\rm \longmapsto \ x^4 + \dfrac{1}{x^4} + 2 = 49$

$\rm \longmapsto \ x^4 + \dfrac{1}{x^4} = 49 - 2$

$\rm \longmapsto \ x^4 + \dfrac{1}{x^4} = 47$

Square on both sides again.

$\rm \longmapsto \ \Bigg( x^4 + \dfrac{1}{x^4} \Bigg)^2 = \big(47\big)^2$

Applying the identity (a + b)² = a² + b² + 2ab we get:

$\rm \longmapsto \ \Big(x^4\Big)^2 + \Bigg(\dfrac{1}{x^4}\Bigg)^2 + 2\Big(x^4\Big)\Bigg(\dfrac{1}{x^4}\Bigg) = 2209$

$\rm \longmapsto \ x^8 + \dfrac{1}{x^8} + 2 = 2209$

$\rm \longmapsto \ x^8 + \dfrac{1}{x^8} = 2209 - 2$

$\rm \longmapsto \ x^8 + \dfrac{1}{x^8} = 2207$

Therefore, the value of x⁸ + (1/x⁸) is 2207.

Answer 2

We've been given that:

$\rm \longmapsto \ x + \dfrac{1}{x} = 3$

Squaring on both sides we get:

$\rm \longmapsto \ \Bigg(x + \dfrac{1}{x} \Bigg)^2 = \big(3\big)^2$

Applying the identity (a + b)² = a² + b² + 2ab we get:

$\rm \longmapsto \ (x)^2 + \Bigg(\dfrac{1}{x} \Bigg)^2 + 2\big(x\big)\Bigg(\dfrac{1}{x} \Bigg) = 9$

$\rm \longmapsto \ x^2 + \dfrac{1}{x^2} + 2 = 9$

$\rm \longmapsto \ x^2 + \dfrac{1}{x^2} = 9 - 2$

$\rm \longmapsto \ x^2 + \dfrac{1}{x^2} = 7$

Square on both sides again.

$\rm \longmapsto \ \Bigg(x^2 + \dfrac{1}{x^2} \Bigg)^2 = \big(7\big)^2$

Applying the identity (a + b)² = a² + b² + 2ab we get:

$\rm \longmapsto \ \Big(x^2\Big)^2 + \Bigg(\dfrac{1}{x^2}\Bigg)^2 + 2\Big(x^2\Big)\Bigg(\dfrac{1}{x^2}\Bigg) = 49$

$\rm \longmapsto \ x^4 + \dfrac{1}{x^4} + 2 = 49$

$\rm \longmapsto \ x^4 + \dfrac{1}{x^4} = 49 - 2$

$\rm \longmapsto \ x^4 + \dfrac{1}{x^4} = 47$

Square on both sides again.

$\rm \longmapsto \ \Bigg( x^4 + \dfrac{1}{x^4} \Bigg)^2 = \big(47\big)^2$

Applying the identity (a + b)² = a² + b² + 2ab we get:

$\rm \longmapsto \ \Big(x^4\Big)^2 + \Bigg(\dfrac{1}{x^4}\Bigg)^2 + 2\Big(x^4\Big)\Bigg(\dfrac{1}{x^4}\Bigg) = 2209$

$\rm \longmapsto \ x^8 + \dfrac{1}{x^8} + 2 = 2209$

$\rm \longmapsto \ x^8 + \dfrac{1}{x^8} = 2209 - 2$

$\rm \longmapsto \ x^8 + \dfrac{1}{x^8} = 2207$

Therefore, the value of x⁸ + (1/x⁸) is 2207.