Question

if a+1/a =5 find a³+1/a³and a²+1/a²​

Answer 1

EXPLANATION.

⇒ (a + 1/a) = 5.

As we know that,

Formula of :

⇒ (x + y)² = x² + y² + 2xy.

⇒ (x + y)³ = x³ + 3x²y + 3xy² + y³.

Using this formula in the equation, we get.

Squaring on both sides of the equation, we get.

⇒ (a + 1/a)² = (5)².

⇒ (a)² + (1/a)² + 2(a)(1/a) = 25.

⇒ a² + 1/a² + 2 = 25.

⇒ a² + 1/a² = 25 - 2.

⇒ a² + 1/a² = 23.

Now, cubing on both sides of the equation, we get.

⇒ (a + 1/a)³ = (5)³.

⇒ (a)³ + 3(a)²(1/a) + 3(a)(1/a)² + (1/a)³ = (5)³.

⇒ a³ + 3a + 3/a + 1/a³ = 125.

⇒ a³ + 3(a + 1/a) + 1/a³ = 125.

Put the value of (a + 1/a = 5) in the equation, we get.

⇒ a³ + 3(5) + 1/a³ = 125.

⇒ a³ + 1/a³ + 15 = 125.

⇒ a³ + 1/a³ = 125 - 15.

⇒ a³ + 1/a³ = 110.

Answer 2

Question:-

$\sf \large If \: a + \frac{1}{a} = 5. \: Find \: a {}^{3} + \frac{1}{a {}^{3} } \: and \: \\ \sf \large a {}^{2} + \frac{1}{a {}^{2} } .$

Given:-

$\sf \large \: a + \frac{1}{a} = 5.$

To Find:-

$\sf \large \: a {}^{3} + \frac{1}{a {}^{3} } \: and \: \sf \large a {}^{2} + \frac{1}{a {}^{2} } .$

Solution:-

We know that,

• (a + b)³ = a³ + b³ + 3ab (a + b).

• a³ + b³ = (a + b)³ – 3ab (a + b).

$\sf \large \underline{1st \: to \: Find \: a {}^{3} + \frac{1}{a {}^{3} } .}$

So,

$\sf \large a {}^{3} + \frac{1}{a {}^{3} } = (a + \frac{1}{a} ) {}^{3} - 3 \times a \times \frac{1}{a} (a + \frac{1}{a} )$

$\sf \large = (5) {}^{3} - 3 \times 5$

$\sf \large= 125 - 15 = 110.$

_______________________________________

$\sf \large \underline{2nd \: to \: Find \:a {}^{2} + \frac{1}{a {}^{2} } . }$

So,

$\sf \large \: a {}^{2} + \frac{1}{a {}^{2} } = (a + \frac{1}{a} ) {}^{2} - 2 \times a \times \frac{1}{a}$

$\sf \large = (5) {}^{2} - 2$

$\sf \large = 25 - 2 = 23.$

Answer:-

$\sf \large \longmapsto \red{ a {}^{3} + \frac{1}{a { }^{3} } = 110.}$

$\sf \large \longmapsto \red{a {}^{2} + \frac{1}{a {}^{2} } = 23.}$

Hope you have satisfied. ⚘