Question

- If x² + 1/ x²= 18, find the value of x³ - 1/x³​

Answer 1

76

Step-by-step explanation:

Given:

${x}^{2} + \dfrac{1}{ {x}^{2} } = 18$

To find:

${x}^{3} - \dfrac{1}{ {x}^{3} }$

Solution:

${x}^{2} + \dfrac{1}{ {x}^{2} } = 18$

[subtracting 2 from both side ]

${x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = 18 - 2 \\ \\ {x}^{2} - 2.x. \frac{1}{x} + \frac{1}{ {x}^{2} } = 16$

[Using a² - 2ab + b² = (a - b)²]

${(x - \frac{1}{ {x}^{} }) }^{2} = 16\\ {(x - \frac{1}{x} )}^{2} = {(4)}^{2} \\ \\ x - \frac{1}{x} = 4 \: \: \: \: ............... \{ i\}$

[cubing both side]

${(x - \dfrac{1}{x}) }^{3} = {(4)}^{3} \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } - 3.x. \frac{1}{x} (x - \frac{1}{x} ) = 64 \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } - 3. \cancel{x}. \frac{1}{\cancel{x}} (4) = 64 \: \: \: \: \: (from \: \: i \: ) \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } - 12 = 64 \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } = 64 + 12 \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } = 76$

Hence, The required answer is 76.

Answer 2

Answer:

here is your answer