A + b = 1 → b = 1 - A
So, subsituting, we have
A2 + (1 - A)2 = 2
A2 + 1 - 2A + A2 = 2 rearrange
2a2 - 2A - 1 = 0 And using the on-site solver and substituting "x" for "A" ......we have....
$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.366\: \!025\: \!403\: \!784\: \!438\: \!6}}\\
{\mathtt{x}} = {\mathtt{1.366\: \!025\: \!403\: \!784\: \!438\: \!6}}\\
\end{array} \right\}$$
So, b = 1 - (-[√3 - 1 ] / 2) = [1 + √3 ] / 2 or b = 1 - [ [1 + √3 ] / 2] = [1 - √3 ] / 2
So
A3 + b3 = ( [1 - √3] / 2 )3 + ( [1 + √3 ] / 2)3 = ( [1 + √3] / 2 )3 + ( [1 - √3 ] / 2)3 = 2.5
