Question

If
$x - \frac{1}{x} = 6$
then, find the value of
${x}^3 - \frac{1}{ {x}^{3} }$

​

Answer 1

Required Answer:-

Given:

• x - 1/x = 6

To find:

• The value of x³ - 1/x³

Solution:

We have,

$\sf \implies x - \dfrac{1}{x} = 6$

We know that,

$\sf \implies {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )$

Therefore,

$\sf \implies {x}^{3} - \dfrac{1}{ {x}^{3} } = { \bigg(x - \dfrac{1}{x} \bigg) }^{3} - 3 \times x \times \dfrac{1}{x} \bigg(x - \dfrac{1}{x} \bigg)$

$\sf \implies {x}^{3} - \dfrac{1}{ {x}^{3} } = { \bigg(x - \dfrac{1}{x} \bigg) }^{3} - 3\bigg(x - \dfrac{1}{x} \bigg)$

Putting the value of x - 1/x, we get,

$\sf \implies {x}^{3} - \dfrac{1}{ {x}^{3} } = {6}^{3} - 3 \times 6$

$\sf \implies {x}^{3} - \dfrac{1}{ {x}^{3} } = 216 - 18$

$\sf \implies {x}^{3} - \dfrac{1}{ {x}^{3} } = 198$

★ Hence, the value of x³ - 1/x³ is 198.

Answer:

• The value of x³ - 1/x³ is 198.

More Identities to know:

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)
• (a + b)² = (a - b)² + 4ab
• (a - b)² = (a + b)² - 4ab
• (a + b)³ = a³ + 3ab(a + b) + b³
• (a - b)³ = a³ - 3ab(a - b) - b³