Question

x^3 +1\x^3=62 then find the valve of root under x ^3+root under 1\x^3​

Answer 1

Answer:

8

Step-by-step explanation:

PLEASE MARK MY ANSWER AS BRAINLIEST

$x^{3} + \frac{1}{x^{3}} = 62$ ......(i)

Using Formulae (a+b)² = a² + b² + 2ab, we get,

$(\sqrt{x^{3} }+\sqrt{\frac{1}{x^{3} } }  )^{2} = x^{3} + \frac{1}{x^{3}} +2*x^{3}*\frac{1}{x^{3} }$ ....(ii)

Now by putting (i) in (ii), we get,

$(\sqrt{x^{3} }+\sqrt{\frac{1}{x^{3} } }  )^{2} = 62 +2$

$\sqrt{x^{3} }+\sqrt{\frac{1}{x^{3} } }   = \sqrt{64}$

$\sqrt{x^{3} }+\sqrt{\frac{1}{x^{3} } }   = 8$