Algebra it

Since, $a, b, c$ are in AP,

$a+c=2b$

$a+b+c=3b$ $\text{(adding b on both the sides)}$

$(a+b+c)^2=3b(a+b+c)$ $\text{(multiplying by a+b+c on both the sides)}$

$(a+b+c)^2=4ab+4bc+3b^2-ab-bc$

$(a+b+c)^2+b(a+c)=4(ab+bc)+3b^2$

By inspection, the solution cannot be the following two options because the degree of the LHS is $\color{#3D99F6}{2}$ while that of the RHS is $\color{#D61F06}{3}$ .

$\begin{aligned} \color{#3D99F6}{(a+b+c)^2+c(a+b)} & = \color{#3D99F6}{4(ab+ac)}+\color{#D61F06}{8c^3} \\ \color{#3D99F6}{a^2+b^2+c^2} & = \color{#3D99F6}{ab+bc+c^2}+\color{#D61F06}{abc} \end{aligned}$

So, I guessed it should be the following:

$\begin{aligned} (a+b+c)^2+b(a+c) & = 4(ab+bc)+3b^2 \quad \quad \small \color{#3D99F6}{\text{For AP, we have }a+c = 2b} \\ (3b)^2 + b(2b) & = 4b(2b) + 3b^2 \\ 9b^2 + 2b^2 & = 8b^2 + 3b^2 \\ \implies 11b^2 & \equiv 11b^2 \end{aligned}$

Therefore, the answer is $\boxed{(a+b+c)^2+b(a+c) = 4(ab+bc)+3b^2 }$ .