Question

9x-9/2x =18 , find the value of x³-1/8x³

Answer 1

$9x - \frac{9}{2x} = 18 \\ \\ = > 9(x - \frac{1}{2x} ) = 18 \\ \\ = > x - \frac{1}{2x} = \frac{18}{9} \\ \\ = > x - \frac{1}{2x} = 2$
$= > (x - \frac{1}{2x} ) {}^{2} = (2) {}^{2}$
${x}^{2} + ( \frac{1}{2x} ) {}^{2} - 2(x)( \frac{1}{x} ) = 4$
$= > {x}^{2} + \frac{1}{4 {x}^{2} } - 1 = 4 \\ \\ = > {x}^{2} + \frac{1}{ {4x}^{2} } = 4 + 1 \\ \\ = > {x}^{2} + \frac{1}{4 {x}^{2} } = 5$

Now,

a³ - b³ = (a - b)(a² + b² + ab)

=>
$= > x {}^{3} - \frac{1}{ {8x}^{3} } = (x - \frac{1}{2x} )( {x}^{2} + \frac{1}{4 {x}^{2} } + x( \frac{1}{2x} ))$

$= > {x}^{3} - \frac{1}{ {8x}^{3} } = (2)(5 + \frac{1}{2} ) \\ \\ = > {x}^{3} - \frac{1}{8 {x}^{3} } = 10 + 1 \\ \\ = > {x}^{3} - \frac{1}{ {8x}^{ 3} } = 11$

Answer 2

Answer:

Step-by-step explanation: