Question

$\huge \bold \red{question - }$

$\huge {x}^{4} + \frac{1}{ {x}^{4} }$
$\huge \: if \: \: \: x + \frac{1}{x} = 7$

$\huge solve \: \: \: the \: \: \: question$
​

Answer 1

$\sf Squaring~ on~ both~ sides:$

$\:$

$\qquad\bf \dashrightarrow \: { \bigg \lgroup x + \dfrac{1}{x} \bigg \rgroup }^{2} = {\bigg \lgroup7 \bigg \rgroup }^{2}$

$\:$

$\bf \qquad \dashrightarrow \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \: \bigg(x \bigg) \: \bigg( \dfrac{1}{x} \bigg) = 49$

$\:$

$\bf \qquad \dashrightarrow \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \: \bigg( \cancel{x} \bigg) \: \bigg( \dfrac{1}{ \cancel{x} } \bigg) = 49$

$\:$

$\qquad \bf \dashrightarrow \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 49 - 2$

$\:$

$\bf \qquad \dashrightarrow \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 47$

$\sf Squaring~ on~ both~ sides~ again:$

$\:$

$\bf \qquad \dashrightarrow \: { \bigg \lgroup {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg \rgroup}^{2} = {\bigg \lgroup47 \bigg \rgroup}^{2}$

$\:$

$\bf \qquad \dashrightarrow \: \bigg( {x ^{2} \bigg)}^{2} + \: \bigg (\dfrac{1}{ {x}^{2} } \bigg) ^{2} + 2 \: \bigg( {x}^{2} \bigg) \: \bigg( \dfrac{1}{ {x}^{2} } \bigg) = 2209$

$\:$

$\bf \qquad \dashrightarrow \: {x}^{4} + \dfrac{1}{ {x}^{4} } + 2 \: \bigg( \cancel{{x}^{2} } \bigg) \: \bigg( \dfrac{1}{ \cancel{{x}^{2} }} \bigg) = 2209$

$\:$

$\bf \qquad \dashrightarrow \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 2209 - 2$

$\:$

$\bf \qquad \dashrightarrow \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 2207$

$\therefore~\sf{The~value~ of~ \bigg\lgroup x^4+\dfrac{1}{x^4}\bigg\rgroup~ is~\pmb{2207}~ when~\bigg\lgroup x+\dfrac{1}{x}\bigg\rgroup~ =7.}$