Question

if a²+1/a²=27 find a+1/a and a³+1/a³​

Answer 1

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

$\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } = 27$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:(i) \: \: a + \dfrac{1}{a} \\ \\ \rm :\longmapsto\:(ii) \: {a}^{3} + \dfrac{1}{ {a}^{3}}$

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

Given that,

$\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } = 27$

Adding 2 on both sides, we get

$\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } + 2= 27 + 2$

$\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 \times a \times \dfrac{1}{a} = 29$

$\rm :\longmapsto\: {\bigg(a + \dfrac{1}{a} \bigg) }^{2} = 29$

$\bf\implies \:a + \dfrac{1}{a} \: = \: \pm \: \sqrt{29}$

Now,

We know,

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} }$

$\rm \:  =  \: {\bigg(a + \dfrac{1}{a} \bigg) }^{3} - 3 \times a \times \dfrac{1}{a} \times \bigg(a + \dfrac{1}{a} \bigg)$

$\rm \:  =  \: {\bigg(a + \dfrac{1}{a} \bigg) }^{3} - 3 \bigg(a + \dfrac{1}{a} \bigg)$

Now,

Case :- 1

$\red{\rm :\longmapsto\:When \: a = \sqrt{29}}$

So,

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } = {( \sqrt{29} )}^{3} - 3( \sqrt{29})$

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } = 29 \sqrt{29} - 3\sqrt{29}$

$\bf :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } = 26\sqrt{29}$

Case :- 2

$\red{\rm :\longmapsto\:When \: a \: = - \: \sqrt{29}}$

So,

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } = {( - \sqrt{29} )}^{3} - 3( - \sqrt{29})$

$\rm :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } = - \: 29 \sqrt{29} + 3\sqrt{29}$

$\bf :\longmapsto\: {a}^{3} + \dfrac{1}{ {a}^{3} } \: = - \: 26\sqrt{29}$

Additional Information :-

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

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

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

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

$\boxed{ \rm \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})}$

$\boxed{ \rm \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})}$

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

Answer 2

$\huge\color{aqua}\ \boxed{\colorbox{black}{Answer : ♞ }}$

$(i) \: {a}^{2} + \frac{1}{ {a}^{2} } = 47$

$Adding \: 2 \: to \: both \: sides, \: we \: get$

$= > {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 47 + 2$

$= > (a + \frac{1}{a} ) = 49$

$Taking \: square \: root \: of \: both \\ sides, \: we \: get :$

$= > \sqrt{(a + \frac{1}{a})^{2} } = \sqrt{49}$

$= > \sqrt{(a + \frac{1}{a}) ^{2} } = \sqrt{7 \times 7}$

$\color{blue} = > a + \frac{1}{a} = 7$

$(ii) \: given :a + \frac{1}{ {a}^{2} } = 7$

$Cubing \: both \: sides, \: we \: get :$

$= > \: (a + \frac{1}{ {a} } ) ^{3} = (7) ^{3}$

$= > (a) ^{3} + ( \frac{1}{a} ) ^{3} + 3 \times a \times \\ \frac{1}{a} (a + \frac{1}{a} ) = 7 \times 7 \times 7$

$= > {a}^{3} + \frac{1}{a ^{3} } + 3(a + \frac{1}{a} ) = 343$

$Putting \: the \: value \: of \:a + \frac{1}{a}, \: we \: get :$

${a}^{3} + \frac{1}{ {a}^{3} } + 3(7) = 343$

$= > a ^{3} + \frac{1}{ {a}^{3} } + 21 = 343$

$= > {a}^{3} + \frac{1}{ {a}^{3} } = 343 - 21$

$= > {a}^{3} + \frac{1}{a ^{3} } = 322$