Question

$if \: x ^{2} + \frac{1}{4x ^{2} } = 8 \: then \: find \: x ^{3} + \frac{1}{8x ^{3} }$<br />mathematics: expansion.<br />don't spam or joke.corono ki kasam.​ <br />slide to view the question.<br />​

Answer 1

Answer:

$\star\: \bf{x^3 + \dfrac{1}{8x^3}}$ = $\sf\blue{\dfrac{45}{2}}$

Given:-

• $\rm x^2 + \dfrac{1}{4x^2} = 8$

To find:-

• $\rm x^3 + \dfrac{1}{8x^3} = ?$

Solution:-

First find the value of $\bold{x + \dfrac{1}{2x}}$

$\rule{100}{1}$

We know that,

$\dag\: \: \sf\blue{a^2 + b^2 = {(a + b)}^{2} - 2ab}$

$\Rightarrow\sf x^2 + \dfrac{1}{4x^2} = 8 \\\\\Rightarrow\sf (x + \dfrac{1}{2x})^2 - 2\times x \times \dfrac{1}{2x} = 8 \\\\\Rightarrow\sf (x + \dfrac{1}{2x})^2 - 1 = 8 \\\\\Rightarrow\sf (x + \dfrac{1}{2x})^2 = 8 + 1 \\\\\Rightarrow\sf (x + \dfrac{1}{2x}) = \sqrt{9} \\\\\Rightarrow{\boxed{\sf\purple{x + \dfrac{1}{2x} = 3 }}}$

$\rule{200}{1}$

We know that,

$\dag\: \: \sf\blue{a^3 + b^3 = (a + b)^3 - 3ab(a + b) }$

$\Rightarrow\sf x^3 + \dfrac{1}{8x^3} = {(x + \dfrac{1}{2x})}^{3} - 3\times x \times \dfrac{1}{2x}(x + \dfrac{1}{2x})$

$\sf \: \: \: \: \: = {3}^{3} - \dfrac{3}{2}(3)$

$\sf \: \: \: \: \: = 27 - \dfrac{9}{2}$

$\sf \: \: \: \: \: = \dfrac{54 - 9}{2}$

$\sf\blue{\: \: \: \: \: = \dfrac{45}{2}}$

$\therefore\bold{Required \: value = \dfrac{45}{2}}$

Answer 2

AnswEr :

$\star$ $\normalsize{\underline{\boxed{\sf \color{violet}{ x^3 + \frac{1}{8x^3} = \frac{45}{2} }}}}$

$\underline{\bigstar\:\textsf{According \: to \: given \: in \: the \: question:}}$

$\normalsize\ : \implies\sf\ x^2 + \frac{1}{4x^2} = 8 \\ \\ \normalsize\ : \implies\sf\ (x + \frac{1}{2x})^2 - 2 \times\ x \times\ \frac{1}{x} = 8$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{Using \: identity, (a+b)^2 = a^2 + b^2 + 2ab}) }$

$\normalsize\ : \implies\sf\ (x + \frac{1}{x})^2 = 9 \\ \\ \normalsize\ : \implies\sf\ x + \frac{1}{x} = \sqrt{9} \\ \\ \normalsize\ : \implies\sf\ x + \frac{1}{x} = 3$

$\normalsize\ : \implies{\boxed{\sf \green{x + \frac{1}{x} = 3 }}}$

$\rule{100}2$

$\bullet$ Using this value find the unknown value.

$\normalsize\ : \implies\sf\ x^3 + \frac{1}{8x^3}$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{Identity, a^3 + b^3 = (a+b)^3 - 3ab(a+b)}) }$

$\normalsize\ : \implies\sf\ (x + \frac{1}{2x})^3 - 3 \times\ x \times\ \frac{1}{x}(x + \frac{1}{2x})$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{Block \: the \: values \: in \: available \: data})}$

$\normalsize\ : \implies\sf\ (3)^3 - \frac{3}{2}(3) \\ \\ \normalsize\ : \implies\sf\ 27 - \frac{9}{2} \\ \\ \normalsize\ : \implies\sf\frac{54-9}{2} = \frac{45}{2}$

$\normalsize\ : \implies{\boxed{\sf \color{darkblue}{x^3 + \frac{1}{8x^3} = \frac{45}{2} }}}$