Value matters #3

$\begin{aligned} a^4 + a^2b^2 + b^4 & = 3 & \small \color{#3D99F6} \text{Given} \\ a^4 + 2a^2b^2 + b^4 & = 3 + a^2b^2 \\ \implies \left(a^2+b^2\right)^2 & = 3 + a^2b^2 & ...(1) \\ a^2 + ab + b^2 & = 3 & \small \color{#3D99F6} \text{Given} \\ a^2 + b^2 & = 3 - ab & \small \color{#3D99F6} \text{Squaring both sides} \\ \implies \left(a^2+b^2\right)^2 & = 9 - 6ab + a^2b^2 & ...(2) \end{aligned}$

Equating $(1)$ and $(2)$ :

$\begin{aligned} 3 + a^2b^2 & = 9 - 6ab + a^2b^2 \\ 6ab & = 6 \\ \implies ab & = 1 \end{aligned}$

From $a^2 + ab + b^2 = 3 \implies a^2 + b^2 = 3 - ab \implies a^2 + b^2 = \boxed{2}$ .

$\begin{aligned} a^4 + a^2 b^2 + b^4 & = (a^2)^2 +2a^2b^2 + (b^2)^2 -a^2b^2 \\ & = (a^2+b^2)^2-(ab)^2 \\ & = (a^2 + b^2 + ab) (a^2 + b^2 - ab) \\ & = (a^2 + ab + b^2 )(a^2-ab +b^2) \\ 3 & = 3(a^2 - ab+b^2) \end{aligned}$

$\implies a^2 - ab+b^2 = 1$

After adding $a^2-ab+b^2 = 1$ and $a^2+ab+b^2 = 3,$ we get $2(a^2+b^2) = 4$

Hence $a^2 + b^2 = \dfrac 42 = \boxed 2$