Question

$If \: x > 0 \: and \: x {}^{2} + \frac{1}{9x {}^{2} } = \frac{25}{36} ,$
So find the value of:
$x {}^{3} + \frac{1}{27 {x}^{3} }$

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Answer 1

$\huge \fbox \green{Answer:}$

Given that:

${x}^{2} + \frac{1}{9 {x}^{2} } = \frac{25}{36}$

$⇒ {x}^{2} + \frac{1}{(3x {)}^{2} } = \frac{25}{36}$

We Know that

$⇒(a + b {)}^{2} = {a}^{2} + 2ab + {b}^{2}$

So Consider the expansion of

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

$⇒(x + \frac{1}{3x} {)}^{2} = {x}^{2} + \frac{2x \times 1}{3x} + \frac{1}{9 {x}^{2} }$

$= {x}^{2} + \frac{2}{3} + \frac{1}{9 {x}^{2} }$

$⇒ {x}^{2} + \frac{1}{(3x {)}^{2} } + \frac{2}{3}$

From

$\sf \colorbox{pink} {we can write that}$

$⇒(x + \frac{1}{3x} {)}^{2} = \frac{25}{36} + \frac{2}{ 3} = \frac{49}{36}$

$∴x + \frac{1}{3x} = \sqrt{ \frac{49}{36} } = \overset{ + }{\longrightarrow} \frac{7}{6}$

Given That ,

$x > 0$

$⇒ \frac{x + 1}{3x} = \frac{7}{6}$

$\text{Now ,}$

$⇒ {x}^{3} + \frac{1}{27 {x}^{3} } = (x + \frac{1}{3x} ) \: \: \: \: \: [ {x}^{2} + \frac{1}{9 {x}^{2} } - \frac{(x)1}{3x}$

$= \frac{7}{6} ( \frac{25}{36} - \frac{1}{3} )$

$= \frac{7}{6} ( \frac{25 - 12}{36} )$

$= \frac{7}{6} \times \frac{13}{36}$

$= \frac{91}{216}$

Hence,

$: {x}^{3} + \frac{1}{27 {x}^{3} } = \frac{91}{216}$

$\\ \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}$

Answer 2

Step-by-step explanation:

hope it helps you ☺️✌️dear