Wir haben zunächst

${\displaystyle {}a_{2}F_{1}-a_{1}F_{2}=(a_{2}b_{1}-a_{1}b_{2})X+a_{2}c_{1}-a_{1}c_{2}\,.}$

Wir ziehen das ${\displaystyle {}a_{1}}$-fache dieses Ausdrucks vom ${\displaystyle {}(a_{2}b_{1}-a_{1}b_{2})}$-fachen von ${\displaystyle {}F_{1}}$ ab und erhalten

${\displaystyle (b_{1}(a_{2}b_{1}-a_{1}b_{2})-a_{1}(a_{2}b_{1}-a_{1}b_{2}))X+(a_{2}b_{1}-a_{1}b_{2})c_{1}-a_{1}(a_{2}c_{1}-a_{1}c_{2}).}$

Indem wir dies wieder über Kreuz multiplizieren erhalten wir

${\displaystyle (b_{1}(a_{2}b_{1}-a_{1}b_{2})-a_{1}(a_{2}b_{1}-a_{1}b_{2}))(a_{2}c_{1}-a_{1}c_{2})-(a_{2}b_{1}-a_{1}b_{2})((a_{2}b_{1}-a_{1}b_{2})c_{1}-a_{1}(a_{2}c_{1}-a_{1}c_{2})).}$

${\displaystyle (a_{2}c_{1}-a_{1}c_{2})^{2}-b_{1}(-a_{2}c_{1}b_{2}-c_{2}a_{2}b_{1}+a_{1}b_{2}c_{2})+c_{1}(a_{1}b_{2}^{2}-2a_{2}b_{1}b_{2}).}$