If a + b + c = 2

\(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\) = 0

ac = \(\frac{4}{b}\)  ,  a³ + b³ + c³ = 28, then

Find a² + b² + c²

8

Formula used → a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)......(i)

If ac = \(\frac{4}{b}\) ⇒ acb = 4

Now; \(\frac{1}{a}\) + \(\frac{1}{b}\) + \(\frac{1}{c}\) = 0

\(\frac{bc + ca + ab}{abc}\) = 0

So, bc + ca + ab = 0

From (i)

28 - 3 (4) = (2) (a² + b² + c² - 0)

16 = 2(a² + b² + c²)

a² + b² + c² = 8