If a + b + c = 2, 1/a+1/b+1/c, ac =4/b and a^3 + b^3 +c^3 = 28, find…

If $a + b + c = 2, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0, ac =\frac{4}{b}$ and $a^3 + b^3 +c^3 = 28$,

find the value of $a^2 + b^2 + c^2$ = ?

We know that,

a³ + b³ + c³ - 3abc = ( a + b + c  ) ( a² + b² + c² - (ab + bc + ca))

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ =0

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

= bc + ac + ab = 0

Given,$ac =\frac{4}{b}$ 

then, abc = 4

Put these values in the given formula,.

28 - 3 × 4 = 2 × (a² + b² + c² - 0)

=  \(\frac{(28 - 12)}{2}\) = a² + b² + c² 

= a² + b² + c² = \(\frac{(16)}{2}\)

= a² + b² + c²  = 8