Question

Please solve in this question​

Answer 1

Question :-

$\rm \: If \: {x}^{2} + \dfrac{4}{ {x}^{2} } = 5, \: then \: find \: {x}^{3} + \dfrac{8}{ {x}^{3} } .$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: {x}^{2} + \dfrac{4}{ {x}^{2} } = 5$

can be rewritten as

$\rm \: {x}^{2} + \dfrac{4}{ {x}^{2} } + 4 = 5 + 4$

$\rm \: {x}^{2} + \dfrac{4}{ {x}^{2} } + 2 \times 2 = 9$

$\rm \: {x}^{2} + \dfrac{4}{ {x}^{2} } + 2 \times x \times \frac{2}{x} = 9$

$\rm \: {x}^{2} + {\bigg(\dfrac{2}{x} \bigg) }^{2} + 2 \times x \times \frac{2}{x} = 9$

We know,

$\boxed{ \rm{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }}$

So, using this identity, we get

$\rm \: {\bigg(x + \dfrac{2}{x} \bigg) }^{2} = 9$

$\bf\implies \:x + \dfrac{2}{x} = \: \pm \: 3$

Now, Consider Case :- 1

$\rm \: When \: \:x + \dfrac{2}{x} = \: 3$

On cubing both sides, we get

$\rm \: \:\bigg(x + \dfrac{2}{x}\bigg)^{3} = \: {3}^{3}$

We know,

$\boxed{ \rm{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }}$

So, using this identity, we get

$\rm \: {x}^{3} + \dfrac{8}{ {x}^{3} } + 3 \times x \times \dfrac{2}{x} \times \bigg(x + \dfrac{2}{x} \bigg) = 27$

$\rm \: {x}^{3} + \dfrac{8}{ {x}^{3} } +6 \times 3 = 27$

$\rm \: {x}^{3} + \dfrac{8}{ {x}^{3} } +18= 27$

$\bf\implies \: {x}^{3} + \dfrac{8}{ {x}^{3} } = 9$

Now, Consider Case : - 2

$\rm \: When \: \:x + \dfrac{2}{x} = \: - 3$

On cubing both sides, we get

$\rm \: \:\bigg(x + \dfrac{2}{x}\bigg)^{3} = \: {( - 3)}^{3}$

We know,

$\boxed{ \rm{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }}$

So, using this identity, we get

$\rm \: {x}^{3} + \dfrac{8}{ {x}^{3} } + 3 \times x \times \dfrac{2}{x} \times \bigg(x + \dfrac{2}{x} \bigg) = - 27$

$\rm \: {x}^{3} + \dfrac{8}{ {x}^{3} } +6 \times ( - 3) = - 27$

$\rm \: {x}^{3} + \dfrac{8}{ {x}^{3} } - 18= - 27$

$\bf\implies \: {x}^{3} + \dfrac{8}{ {x}^{3} } = - \: 9$

So, from above we concluded that

$\bf\implies \: {x}^{3} + \dfrac{8}{ {x}^{3} } = \pm \: 9$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

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$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

$\rule{300pt}{0.1pt}$

$\begin{array}{ll} \rm \underline{\underline{\text { Sol }^{n}}} & \displaystyle \rm x^{2}+\frac{4}{x^{2}}=5 \\ \\ &\displaystyle \rm \left(x+\frac{2}{x}\right)^{2}-2 \times x \times \frac{2}{x}=5 \\ \\ &\displaystyle \rm \left(x+\frac{2}{x}\right)^{2}-4=5 \\ \\ & \displaystyle \rm\left(x+\frac{2}{x}\right)^{2}=9 \\ \\ &\displaystyle \rm\left(x+\frac{2}{x}\right)^{2}=(3)^{2} \\ \\ \Rightarrow &\displaystyle \rm x+\frac{2}{x}=3\end{array}$

Now; $\displaystyle \rm \left(x+\frac{2}{x}\right)^{3}=x^{3}+\left(\frac{2}{x}\right)^{3}+3 \times x \times \frac{2}{x}\left(x+\frac{2}{x}\right)$

$\[ \begin{array}{l} \displaystyle\rm \therefore x^{3}+\left(\frac{2}{x}\right)^{3}=\left(x+\frac{2}{x}\right)^{3}-3 \times x \times \frac{2}{x}\left(x+\frac{2}{x}\right) \\ \\ \displaystyle\rm x^{3}+\frac{8}{x^{3}}=(3)^{3}-3 \times 2(3) \\\\ \displaystyle\rm x^{3}+\frac{8}{x^{3}}=27-18 \\\\ \color{darkorange} \boxed{ \displaystyle\rm \therefore x^{3}+\frac{8}{x^{3}}=9} \end{array} \]$

Answer :-Value of $\rm x^{3}+\dfrac{8}{x^{3}}$ is 9