Question

if x^2+ 1/x^2 =43 find the value of x^3- 1/x^3

Answer 1

In the above Question , the following information is given -

x² + 1 / x² = 43 .

To Find -

x³ - 1 / x³

Solution -

x² + 1 / x² = 43 .

=> x² + 1 / x² - 2 = 43 - 2

=> x² + 1 / x² - 2 = 41

=> ( x - 1 / x ) ² = 41

=> x - 1 / x = 41½

Cubing -

(x - 1 / x ) ³ = 41^ { 3 / 2 }

=> x³ - 1 / x³ - 3 ( x )( 1 / x )( x - 1 / x ) = 41^ { 3 / 2 }

=> x³ - 1 / x³ - 3 × 41½ = 41 ^ { 3 / 2 }

=> x³ - 1 / x³ = 41 ^ { 3 / 2 } - 41½

This is the answer ...

_____________

Answer 2

$\huge\sf\blue{Given}$

✭ $\sf x^2 + \dfrac{1}{x^2} = 43$

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

☆ $\sf x^3 - \dfrac{1}{x^3}$

$\rule{110}1$

$\huge\sf\purple{Steps}$

So now,

➝ $\sf x^2 + \dfrac{1}{x^2} - 2 = 43 - 2$

➝ $\sf x^2 + \dfrac{1}{x^2} - 2 = 41$

➝ $\sf (x^2 - \dfrac{1}{x^2})^2 = 41$

➝ $\sf (x^2 - \dfrac{1}{x^2} = {41}\dfrac{1}{2}$

On cubing,

➢ $\sf (x - \dfrac{1}{x}^3) = 41^\frac{3}{2}$

➢ $\sf x^3 - \dfrac{1}{x^3} - 3(x)(\dfrac{1}{x})(x- \dfrac{1}{x}) = 41^\frac{3}{2}$

➢ $\sf x^3 - \dfrac{1}{x^3} = 4{1}^{\frac{3}{2}} + 41 \dfrac{1}{2}$

➢ $\sf\orange{x^3 - \dfrac{1}{x^3} = 41^\frac{3}{2} - 41 \dfrac{1}{2}}$

$\rule{170}3$