Question

if [a+1/a] square =9 [a cube +1/a cube] equals to​

Answer 1

EXPLANATION.

⇒ (a + 1/a)² = 9.

As we know that,

We can write equation as,

⇒ (a + 1/a) = √9.

⇒ (a + 1/a) = 3.

Cubing on both sides of the equation, we get.

⇒ (a + 1/a)³ = (3)³.

As we know that,

Formula of :

⇒ (x + y)³ = x³ + 3x²y + 3xy² + y³.

Using this formula in the equation, we get.

⇒ (a)³ + 3(a)²(1/a) + 3(a)(1/a)² + (1/a)³ = (3)³.

⇒ a³ + 3a + 3/a + 1/a³ = 27.

⇒ a³ + 3(a + 1/a) + 1/a³ = 27.

Put the value of (a + 1/a) = 3 in the equation, we get.

⇒ a³ + 3(3) + 1/a³ = 27.

⇒ a³ + 1/a³ + 9 = 27.

⇒ a³ + 1/a³ = 27 - 9.

⇒ a³ + 1/a³ = 18.

Answer 2

$\underline{ \bf SOLUTIO} :$

$\rm⇢ \: \: { [a + \frac{1}{a} ]}^{2} = 9$

$\rm⇢ \: \: a + \frac{1}{a} = \sqrt{9} = 3$

from this by Cubing on Both the sides,

$\rm⇢ \: \: [a + \frac{1}{a} ]^{3} = 27$

then, By using

(a + b)³ = a³ + b³ + 3a²b + 3b²a , we get

$\rm⇢ \: \: {a}^{3} + \frac{1}{ {a}^{3} } + 3 {(a)}^{ \cancel2} ( \frac{1}{ \cancel a} ) + 3( \cancel a)( \frac{1}{ {a}^{ \cancel2} } ) = 27$

$\rm⇢ \: \: {a}^{3} + \frac{1}{ {a}^{3} } + 3a + \frac{3}{a} = 27$

$\rm⇢ \: \: 3(a + \frac{1}{a} ) + {a}^{3} + \frac{1}{ {a}^{3} }$

Substitute a + 1/a = 3, we get

$\rm⇢ \: \: 3(3) + {a}^{3} + \frac{1}{ {a}^{3} } = 27$

$\rm⇢ \: \: {a}^{3} + \frac{1}{ {a}^{3} } = 27 - 9 = 18$

FINAL ANSWER :

$\longmapsto \: \: \bf {a}^{3} + \frac{1}{ {a}^{3} } = 18$