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
Answer & explanation
Correct answer: option 2
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