An algebra problem by durgaprasad sahu

$\begin{aligned} \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \\ & = \frac{1}{3-a}+\frac{1}{3-b}+\frac{1}{3-c} = \frac{10}{3} \\ & = \frac{(3-b)(3-c)+(3-a)(3-c)+(3-a)(3-b)}{(3-a)(3-b)(3-c)}=\frac{10}{3} \\ & \implies 3\times \left(27-6(a+b+c)+ab+bc+ca\right)=10 \times (3-a)(3-b)(3-c) \\ & \implies 81-18(a+b+c)+3(ab+bc+ca)=270-90(a+b+c)+30(ab+bc+ca)-10abc \\ & \implies 189-72(a+b+c)+27(ab+bc+ca)=10abc \\ & \implies 21-8(\color{#D61F06}{\underbrace{a+b+c}_3})\color{#333333}+3(ab+bc+ca)=\frac{10}{9}abc \\ & = 21-24+3(ab+bc+ca) = \frac{10}{9}abc \\ & \implies \frac{ab+bc+ca-1}{abc} = \frac{10}{27} \end{aligned}$

Also, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=\frac{ab+bc+ca-1}{abc}=\boxed{\dfrac{10}{27}}$

$\begin{aligned} \frac 1{a+b} + \frac 1{b+c} + \frac 1{c+a} & = \frac {10}3 & \small \color{#3D99F6} \text{Since }a+b+c = 3 \\ \frac 1{3-c} + \frac 1{3-a} + \frac 1{3-b} & = \frac {10}3 \\ \frac {(3-a)(3-b)+(3-b)(3-c)+(3-c)(3-a)}{(3-a)(3-b)(3-c)} & = \frac {10}3 \\ \frac {27-6{\color{#3D99F6}(a+b+c)}+ab+bc+ca}{27-9{\color{#3D99F6}(a+b+c)}+3(ab+bc+ca)-abc} & = \frac {10}3 & \small \color{#3D99F6} \text{Again }a+b+c = 3 \\ \frac {9+ab+bc+ca}{3(ab+bc+ca)-abc} & = \frac {10}3 \\ 27 + 3(ab+bc+ca) & = 30(ab+bc+ca)-10abc \\ 27(ab+bc+ca) - 27 & = 10abc & \small \color{#3D99F6} \text{Divide both sides by }27abc \\ \frac 1a+\frac 1b + \frac 1c - \frac 1{abc} & = \boxed{\dfrac {10}{27}} \end{aligned}$