Question

solve question number b I you can ​

Answer 1

Challenge Accepted!

$\frac{ {a}^{2} + 1 }{a} = 4$

$= > \frac{ {a}^{2} }{a} + \frac{1}{a} = 4$

$= > a + \frac{1}{a} = 4$

$= > {(a + \frac{1}{a}) }^{3} = {4}^{3}$

$= > {a}^{3} + \frac{1}{ {a}^{3} } + 3a \times \frac{1}{a}(a + \frac{1}{a} ) = 64$

$= > {a}^{3} + \frac{1}{ {a}^{3} } + 3 \times 1 \times 4 = 64$

$= > {a}^{3} + \frac{1}{ {a}^{3} } + 12 = 64$

$= > {a}^{3} + \frac{1}{ {a}^{3} } = 64 - 12 = 52$

$= > 2( {a}^{3} + \frac{1}{ {a}^{3} } ) = 52 \times 2$

$= > 2 {a}^{3} + \frac{2}{ {a}^{3} } = 104$

__________

HOPE IT HELPS❤️