If 8K⁶ + 15K³ - 2 = 0

Find K + \(\frac{1}{K}\) (positive value)

2\(\frac{1}{2}\)

Formula → 15K³ = 16K³ - K³ 

8K⁶ + 16K³ - K³ - 2 = 0

8K³ (K³ + 2) - 1 (K³ + 2)= 0

K³ = \(\frac{1}{8}\) or K³ = -2

So if, K³ = \(\frac{1}{8}\) → K = \(\sqrt [3]{\frac{1}{8}}\) = \(\frac{1}{2}\)

⇒ K + \(\frac{1}{K}\) = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\) or 2\(\frac{1}{2}\)