An algebra problem by me...

$a^{2}+b^{2}+c^{2} = (a+b+c)^2 - 2(ab+bc+ca)=9$

$putting$ $values$ $in$ $this,$

$1 - 2(ab+bc+ca) = 9$

$ab+bc+ca = -4------------(1)$

$now$

$(a^{2}+b^{2}+c^{2})(a+b+c) = a^{3}+b^{3}+c^{3} + ab(a+b) + bc(b+c) +$

$ca(c+a) = 9$

$1 + ab(1-c) + bc(1-a) + ca(1-b) = 9$

$ab + bc + ca - 3abc = 8$

$abc = -4----------------(2)$

$modifying$ $the$ $expression$ $we$ $need$ $to$ $find,$

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + bc + ca}{abc}$

$putting$ $(1)$ $and$ $(2)$ , $we$ $get$

$\frac{-4}{-4} = \boxed{1}$

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

$\Rightarrow$ $\frac {ab+bc+ca}{abc}$ ---->(1)

We know that

$(a+b+c)^2$ = $a^2$ + $b^2$ + $c^2$ +2(ab+bc+ca)

By substituting the given values in the above identity, we get

2(ab+bc+ca)=1-9

$\Rightarrow$ ab+bc+ca=-4---->(2)

We know that

$a^3$ + $b^3$ + $c^3$ =(a+b+c)( $a^2$ + $b^2$ + $c^2$ -(ab+bc+ca))+3abc

By substituting the given values and (1) in the above identity, we get

1=(1)(9-(-4))+3abc

$\Rightarrow$ 3abc=-12

$\Rightarrow$ abc=-4---->(3)

(1)= $\frac {(2)}{(3)}$

$\Rightarrow$ $\frac {(-4)}{(-4)}$ = $\boxed{1}$

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca So value of ab+bc+ca is -4

And from a^3+b^3+c^3=(a+b+c+)(a^2+b^2+c^3) + 3abc And value of abc come -4 1/a+1/b+1/c= (ab+bc+ca)/abc =-4/-4 U gotta answer and I gotta nothing :-( Nice question Sarthak ghoda gadi.......:-P