Question

if
$a + 1 \div a = 8 \: find \: the \: value \: of {a}^{6 } + \frac{1}{ {a}^{6} }$
​

Answer 1

$\large \bf{}Given : \boxed{ \tt \: \: a + \frac{1}{a} = 8 \: }$

$\rm{}To \: \: find : \boxed{ \bf \: \: \small {{x}^{6} + \frac{1}{ {x}^{6} } = \: ? } }$

$\huge{ : \mathbb{SOLUTION : }}$

Here,

$\rm{}a + \frac{1}{a} = 8 \\ \small\boxed{ \rm{}squaring \: \: both \: \: sides} \\ \to\rm { \bigg(a + \frac{1}{a} \bigg)}^{2} = {(8)}^{2}$

• Identity Required : (a+b)² = a² +2ab+ b²

$\to \rm {a}^{2} + \frac{1}{ {a}^{2} } + 2(\cancel{a}) \bigg( \frac{1}{ \cancel{a}} \bigg) = 64 \\ \\ \to \rm {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 64 \\ \small \boxed{ \rm{transposing \: \: 2 \: \: to \: \:RHS }} \\ \to \rm {a}^{2} + \frac{1}{ {a}^{2} } = 64 - 2 \\ \\ \to \overline{ \underline{ \underline{ \rm{ {a}^{2} + \frac{1}{ {a}^{2} } = 62}}}}$

Now,

On Cubing both sides of obtained equation, we get,

$\rm {\bigg( {a}^{2} + \frac{ 1}{ {a}^{2} } \bigg)}^{3} = {(62)}^{3}$

• Identity : (a+b)³ = a³ + b³ + 3ab(a+b)

Using above identity, we get,

$\small \to \rm {( {a}^{2} )}^{3} + { \bigg( \frac{1}{ {a}^{2} } \bigg)}^{3} + 3( \cancel{{a}^{2}} )\bigg( \frac{1}{ \cancel{{a}^{2}} } \bigg) \bigg\{ {a}^{2} + \frac{1}{ {a}^{2} } \bigg\} = 238328 \\ \\ \small \to \rm {a}^{6} + \frac{1}{ {a}^{6} } + 3 \bigg({a}^{2} + \frac{1}{ {a}^{2}} \bigg) = 238328 \\ \\ \small \to \rm {a}^{6} + \frac{1}{ {a}^{6} } + 3(62) = 238328\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{ \because \: {a}^{2} + \frac{1}{ {a}^{2} } = 62 \} \\ \\ \to \rm{a}^{6} + \frac{1}{ {a}^{6} } + 186= 238328 \\ \small \boxed{ \rm{transposing \: \: 186 \: \: to \: \: RHS}} \\ \to \rm{a}^{6} + \frac{1}{ {a}^{6} } = 238328 - 186 \\ \\ \to \rm{a}^{6} + \frac{1}{ {a}^{6} } = \red{238142}$

Hence,

• Value of x⁶ + 1/x⁶ = 238142

Some Important Formulae :

➳ ( x - y )² = x² - 2xy + y²

➳ ( x - y ) ( x -y ) = ( x - y )²

➳ (x - y)³ = x³ - y³ - 3xy(x - y)

➳ ( x + y ) ( x + y ) = ( x + y )²

➳ x² - y² = ( x + y ) ( x - y )

➳ ( x + a ) ( x + b ) = x² + ( a + b)x + ab