Question

If a² + (1/a²) = 23 and a ≠ 0, find the value of a³ + (1/a³)

Answer 1

The value of $a^{3}+\frac{1}{a^{3} }$ is 110

• Given

              $a^{2}+\frac{1}{a^{2} }=23$

• We know that

            $(a+b)^{2}=a^{2}+b^{2}+2ab\\(a+\frac{1}{a})^{2}=a^{2}+\frac{1}{a^{2} } +2\\(a+\frac{1}{a})^{2}=23+2\\(a+\frac{1}{a})^{2}=25\\a+\frac{1}{a}=5\\Now,\\a^{3}+\frac{1}{a^{3}}=(a+\frac{1}{a})(a^{2} +\frac{1}{a^{2} }-a(\frac{1}{a}))\\ a^{3}+\frac{1}{a^{3}}=(5)(23-1)\\a^{3}+\frac{1}{a^{3}}=5(22)\\a^{3}+\frac{1}{a^{3}}=110$

Answer 2

a³ +  $\frac{1}{a^{3} }$ = 110

Explanation:

Given,

$a^{2}$ + $\frac{1}{a^{2} }$ = 23

Applying the formula: (a + b)² = a² + b² + 2ab

a² + b² =  (a + b)² - 2ab

Putting the given value:

23 = (a + $\frac{1}{a}$)² - 2 × a ×  $\frac{1}{a}$

23 = (a + $\frac{1}{a}$)² - 2

23 + 2 = (a + $\frac{1}{a}$)²

(a + $\frac{1}{a}$) = $\sqrt{25}$

(a + $\frac{1}{a}$) = 5

To find: a³ + $\frac{1}{a^{3} }$

Applying the formula: (a + b)³ = a³ + b³ + 3a²b + 3ab²

(a + $\frac{1}{a}$)³ = a³ +  $\frac{1}{a^{3} }$ + 3 × a² ×  $\frac{1}{a}$ + 3 × a ×  $\frac{1}{a^{2} }$

(5)³ = a³ +  $\frac{1}{a^{3} }$ + 3a + $\frac{3}{a}$

125 = a³ +  $\frac{1}{a^{3} }$ + 3 (a + $\frac{1}{a}$)

125 = a³ +  $\frac{1}{a^{3} }$ + 3 (5)

125 = a³ +  $\frac{1}{a^{3} }$ + 15

a³ +  $\frac{1}{a^{3} }$ = 125 - 15

a³ +  $\frac{1}{a^{3} }$ = 110

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