The equation $(a^2 + b^2 - 1)(a + b) = (b^2 + c^2 - 1)(b + c)$ is \begin{align*} a^3 + a^2 b + a b^2 + b^3 - (a + b) & = c^3 + c^2 b + c b^2 + b^3 - (c + b)\\ \iff (a^3 - c^3) + (a^2 - c^2) b + (a - c) b^2 - (a - c) & = 0\\ \iff (a - c) \big[a^2 + ac + c^2 + (a + c) b + b^2 - 1\big] & = 0\\ \iff a^2 + b^2 + c^2 + ab + bc + ca & = 1 \end{align*}as $a \ne c$. Similarly $d^2 + b^2 + c^2 + db + bc + cd = 1$, so subtracting these two, we get \begin{align*} (a^2 - d^2) + (a - d)b + (a - d)c & = 0\\ \iff (a - d) \big[a + d + b + c\big] & = 0\\ \iff a + b + c + d & = 0 \end{align*}as $a \ne d$.