Question

if x2 + 1/x2 =11 then find the value of x3 -1/x3.

Answer 1

Given ⇒
x² + 1/x² = 11

Now, Using the Formula,
(a - 1/a)² = a² + 1/a² - 2
∴ (x - 1/x)² = x² + 1/x² - 2
∴ (x - 1/x)² = 11 - 2
∴ x - 1/x = √9
⇒ x - 1/x = 3

Now,
Using the Formula,
(a - 1/a)³ = a³ - 1/a³ - 3(a - 1/a)
∴ (x - 1/x)³ = x³- 1/x³ - 3(x - 1/x)
∴ (3)³ = x³ - 1/x³ - 3(3)
⇒ 27 = x³ - 1/x³ - 9
⇒ x³ - 1/x³ = 27 + 9
⇒ x³ - 1/x³ = 36

Hence, the value of the x³ - 1/x³ is 36.


Hope it helps.

Answer 2

Final Answer :±36

Steps :
1)
${x}^{2} + \frac{1}{ {x}^{2} } = 11 \\ = > {(x - \frac{1}{x} )}^{2} + 2 = 11 \\ = > {(x - \frac{1}{x} )}^{2} = 11 - 2 = 9 \\ = > (x - \frac{1}{x} ) = + 3 \: \: or \: \: - 3$

2) Case : 1
When
$x - \frac{1}{x} = 3$
then,
${x}^{3} - \frac{1}{ {x}^{3} } = {(x - \frac{1}{x})}^{3} + 3(x - \frac{1}{x} ) \\ = > {3}^{3} + 3 \times 3 \\ = > 27 + 9 = 36$

3) Case :2
When
$x - \frac{1}{x} = - 3$
then,
${x}^{3} - \frac{1}{ {x}^{3} } = {( - 3)}^{3} + 3 \times ( - 3) \\ = > - 27 - 9 \\ = > - 36$

So,
x^3 - 1/x^3 = ±36