Question

please koi bahana mat do please ye bata do 1 last
a humble request​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

→ a = 3 + √8

→ 1/a = 1/(3 + √8)

→ 1/a = 1/(3 + √8) * (3 - √8) / (3 - √8)

→ 1/a = (3 - √8)/((3 + √8)(3 - √8)

→ 1/a = (3 - √8)/(3² - (√8)²)

→ 1/a = (3 - √8)/(9 - 8)

→ 1/a = (3 - √8)

So,

→ (a + 1/a) = 3 + √8 + 3 - √8 = 6 (Ans.)

_________________

Now,

→ (a + 1/a) = 6

Squaring both sides,

→ (a + 1/a)² = 6²

using (a + b)² = a² + b² + 2ab in LHS,

→ a² + 1/a² + 2 * a * 1/a = 36

→ a² + 1/a² + 2 = 36

→ (a² + 1/a²) = 36 - 2

→ (a² + 1/a²) = 34 (Ans.)

_________________

Now,

→ (a² + 1/a²) = 34

Squaring both sides,

→ (a² + 1/a²)² = 34²

using (a + b)² = a² + b² + 2ab in LHS,

→ a⁴ + 1/a⁴ + 2 * a² * 1/a² = 1156

→ a⁴ + 1/a⁴ + 2 = 1156

→ (a⁴ + 1/a⁴) = 1156 - 2

→ (a⁴ + 1/a⁴) = 1154 (Ans.)

_________________

And, in Last,

→ (a + 1/a) = 6

cubing both sides, we get,

→ (a + 1/a)³ = 6³

using (a + b)³ = a³ + b³ + 3ab(a + b) in LHS, we get,

→ a³ + 1/a³ + 3 * a * 1/a(a + 1/a) = 216

→ a³ + 1/a³ + 3(a + 1/a) = 216

Putting value of (a + 1/a) = 6 Now,

→ a³ + 1/a³ + 3*6 = 216

→ a³ + 1/a³ + 18 = 216

→ (a³ + 1/a³) = 216 - 18

→ (a³ + 1/a³) = 198 (Ans.)

___________________________

Answer 2

${ \huge{ \bold{ \mathfrak{ \underline{ \red{Question:-}}}}}}$

▪ If

$\bold{a = 3 + \sqrt{8}}$

find

${ \bold{ \: \: a + \frac{1}{a} }} \\ \\ \:{ \bold{ {a}^{2} + \frac{1}{ {a}^{2} } }}$

${ \bold{ {a}^{3} + \frac{1}{ {a}^{3} } }} \\ \\ { \bold{ {a}^{4} + \frac{1}{ {a}^{4} } }}$

${ \huge{ \underline{ \mathfrak{ \red{Solution:-}}}}}$

▪ it's given in the question that....

${ \bold{ a = 3 + \sqrt{8} }}$

then,

${ \bold{ \frac{1}{a} = \frac{1}{3 + \sqrt{8} } }}$

▪ rationalizing the denominator...

multiplying 3-root 8 to both numerator and denominator....

${ \bold{ \implies{ \frac{1}{a} = \frac{1}{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{3 - \sqrt{8}}}} }$

${ \bold{ \implies{ \frac{1}{a} = \frac{3 - \sqrt{8}}{(3 + \sqrt{8})(3 - \sqrt{8})}} } }$

▪ using the algebraic identity in the denominator....

${ \boxed{ \bold{ \red{ (a + b)(a - b) = {a}^{2} - {b}^{2} }}}}$

${ \bold{ \implies{ \frac{1}{a} = \frac{3 - \sqrt{8} }{ {3}^{2} - {( \sqrt{8} )}^{2}}}} }$

${ \bold{ \implies{ \frac{1}{a} = \frac{3 - \sqrt{8} }{(9 - 8)}}}}$

${ \bold{ \implies{ \blue{ \frac{1}{a} = {3 - \sqrt{8} }}}}}$

thus,

${ \bold{ \red{ \implies{a + \frac{1}{a} }}} }{ \bold{ =(3 + \sqrt{8} ) + (3 - \sqrt{8}) }}$

${ \underline{ \boxed{ \bold{ \red{ \implies{a + \frac{1}{a} }} { \bold{= 3 + 3 = 6}}}}}}$

${ \bold{ a + \frac{1}{a} = 6}}$

▪ squaring both the sides....

${ \bold{ \implies{ {(a + \frac{1}{a}) }^{2} = {6}^{2} }}}$

▪ using. the algebraic identity given below in L.H.S....

${ \bold{ \underline{ \pink{ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}}}}$

${ \bold{ \implies{ {a}^{2} + \frac{1}{ {a}^{2} } + 2 \times a \times \frac{1}{a} = 36}}}$

${ \bold{ \implies{ {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 36}}}$

${ \underline {\boxed{ \bold{ \red{ \implies{ {a}^{2} + \frac{1}{ {a}^{2} }}}}{ \bold{ = 36 - 2 = 34}}}}}$

${ \bold{ {a}^{2} + \frac{1}{ {a}^{2} } = 34}}$

▪ again squaring both the sides...

${ \bold{ \implies{ {( {a}^{2} + \frac{1}{ {a}^{2} } )}^{2} = {(34)}^{2} }}}$

${ \bold{ \underline{ \pink{ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}}}}$

▪ using the above formula again in L.H.S ....

${ \bold{ \implies{ {a}^{4} + \frac{1}{ {a}^{4} } + 2 \times {a}^{2} \times \frac{1}{ {a}^{2} } = 1156}}}$

${ \bold{ \implies{ {a}^{4} + \frac{1}{ {a}^{4} } + 2 = 1156}}}$

${ \underline{ \boxed{ \bold{ \red{ \implies{ {a}^{4} + \frac{1}{ {a}^{4} }}}}{ \bold{ = 1156 - 2 = 1154}}} }}$

${ \bold{a + \frac{1}{a} = 6}}$

▪ cubing both the sides....

${ \bold{ \implies{ {(a + \frac{1}{a} )}^{3} = {6}^{3}}}}$

▪ now , using the given formula in L.H.S..

${ \bold{ \underline{ \pink{ {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)}}}}$

${ \bold{ \implies{ {a}^{3} + \frac{1}{ {a}^{3} } + 3a \times \frac{1}{a} (a + \frac{1}{a} ) = 216}}}$

${ \bold{ \implies{ {a}^{3} + \frac{1}{ {a}^{3} } + 3(a + \frac{1}{a} ) = 216}}}$

▪ putting the value of (a+1/a)=6, in the above equation...

${ \bold{ \implies{ {a}^{3} + \frac{1}{ {a}^{3} } + 3(6) = 216}}}$

${ \bold{ \implies{ {a}^{3} + \frac{1}{ {a}^{3} } + 18 = 216}}}$

${ \underline{ \boxed{ \bold{ \red{ \implies{ {a}^{3} + \frac{1}{ {a}^{3} } }}}{ \bold{ = 216 - 18 = 198}}}}}$