Question

If y² + 1/y² = 14, then find the value of y³ + 1/y³

Answer 1

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

Given that

$\red{\rm :\longmapsto\: {y}^{2} + \dfrac{1}{ {y}^{2} } = 14}$

On adding 2 on both sides,we get

${\rm :\longmapsto\: {y}^{2} + \dfrac{1}{ {y}^{2} } + 2= 14 + 2}$

can be rewritten as

${\rm :\longmapsto\: {y}^{2} + \dfrac{1}{ {y}^{2} } + 2 \times y \times \dfrac{1}{y} = 16}$

We know that

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

So, using this identity, we get

$\rm :\longmapsto\: {\bigg(y + \dfrac{1}{y} \bigg) }^{2} = 16$

$\rm :\longmapsto\: y + \dfrac{1}{y} = \pm \: 4$

Two cases arises.

Consider Case - 1

$\rm :\longmapsto\: y + \dfrac{1}{y} = \: 4$

On cubing both sides, we get

$\rm :\longmapsto\: {\bigg(y + \dfrac{1}{y} \bigg) }^{3} = {4}^{3}$

$\rm :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } + 3 \times y \times \dfrac{1}{y}\bigg(y + \dfrac{1}{y} \bigg) = 64$

$\rm :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } + 3 \times 4 = 64$

$\rm :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } + 12 = 64$

$\rm :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } = 64 - 12$

$\bf :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } = 52$

Consider Case - 2

$\rm :\longmapsto\: y + \dfrac{1}{y} = \: - \: 4$

On Cubing both sides, we get

$\rm :\longmapsto\: {\bigg(y + \dfrac{1}{y} \bigg) }^{3} = {( - 4)}^{3}$

$\rm :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } + 3 \times y \times \dfrac{1}{y}\bigg(y + \dfrac{1}{y} \bigg) = - 64$

$\rm :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } + 3 \times ( - 4 ) = - 64$

$\rm :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } - 12 = - 64$

$\rm :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } = - 64 + 12$

$\bf :\longmapsto\: {y}^{3} + \dfrac{1}{ {y}^{3} } = - 52$

Additional Information :-

More Identities to know

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

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

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

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

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

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

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

$\boxed{ \rm{ {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2})}}$

Answer 2

Answer:

Given, y² + 1/y² = 23

⇒ (y + 1/y)² - (2 × y × 1/y) = 23

⇒ (y + 1/y)² - 2 = 23

⇒ (y + 1/y)² = 23 + 2

⇒ (y + 1/y)² = 25

⇒ y + 1/y = - 5,

where we take the negative value only.

∴ y³ + 1/y³

= (y + 1/y)³ - (3 × y × 1/y)(y + 1/y)

= (- 5)³ - {3 × (- 5)}

= - 125 + 15

= - 110

Identity rule :

a³ + b³ = (a + b)³ - 3ab(a + b)

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