Question

if X square + 1 by 4 x square equals to 15 find the value of x cube + 1 by 8 x cube​

Answer 1

Answer:

58

Step-by-step explanation:

Given :

If,

$x^2 + \frac{1}{4x^2} = 15$ ...(i)

Then,

To find the value of :

$x^3 + \frac{1}{8x^3}$

Solution :

We know that,

a³ + b³ = (a + b) (a² - ab + b²)    ..(ii)

&

(a + b)² = a² + 2ab + b²  ..(iii)

We also knew that,

$x^2 + \frac{1}{4x^2} = 15$

⇒ $x^2 + \frac{1}{4x^2} + 1 = 15 + 1$

⇒ $(x)^2 + (\frac{1}{2x})^2 + 2(x)(\frac{1}{2x}) = 16$

⇒ $(x + \frac{1}{2x})^2 = 16$

By taking square roots on both the sides,

We get,

⇒ $x + \frac{1}{2x} = 4$ ...(iv)

_

$(x)^3 + (\frac{1}{2x})^3 = (x + \frac{1}{2x})(x^2 - (x)(\frac{1}{2x}) + (\frac{1}{2x}) ^2)$ by (ii)

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

from (i) , (iii) & (iv),

We get,

$x^3 + \frac{1}{8x^3} = (4)(15 - \frac{1}{2}) = 58$