If $\mathrm{a}, \mathrm{b}\in \mathrm{R}$ such that a + b = 1 and (1 - 2ab)(a³ + b³) = 12. The value of (a² + b²) is equal to ________.

Let ${\mathrm{a}}^2+{\mathrm{b}}^2=\mathrm{x}$

$1-2\mathrm{ab}=(\mathrm{a}+\mathrm{b}{)}^2-2\mathrm{ab}={\mathrm{a}}^2+{\mathrm{b}}^2=\mathrm{x}$

$\therefore \mathrm{ab}=\frac{1-\mathrm{x}}{2}$

and ${\mathrm{a}}^3+{\mathrm{b}}^3=(\mathrm{a}+\mathrm{b})\left({\mathrm{a}}^2+{\mathrm{b}}^2-\mathrm{ab}\right)$

${\mathrm{a}}^3+{\mathrm{b}}^3=\mathrm{x}-\mathrm{ab}$                …… (1)

but $\mathrm{ab}=\frac{1-\mathrm{x}}{2}$

Hence ${\mathrm{a}}^3+{\mathrm{b}}^3=\mathrm{x}-\frac{1-\mathrm{x}}{2}=\frac{3\mathrm{x}-1}{2}$

Hence the equation

$(1-2\mathrm{ab})\left({\mathrm{a}}^3+{\mathrm{b}}^3\right)=12$, becomes

$\mathrm{x}\left(\frac{3\mathrm{x}-1}{2}\right)=12$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3{\mathrm{x}}^2-\mathrm{x}-24=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{x}-3)(3\mathrm{x}+8)=0\end{array}$

$\Rightarrow \mathrm{x}=3\text{ or }\mathrm{x}=-\frac{8}{3}$ (not possible) as $\mathrm{x}={\mathrm{a}}^2+{\mathrm{b}}^2<0$

$\therefore \mathrm{x}=3\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=3$