We know $\angle AMB=\angle ANB = 90$ . So by Pythagoras theorem, $AB^2=AM^2+BM^2 \implies x^2=a^2+BM^2 \implies BM=\sqrt{x^2-a^2}$. Simillarly $AN=\sqrt{x^2-c^2}$. Now by Ptolemy's theorem , $AB.MN+AM.BN=AN.BM $ $\implies xb+ac=\sqrt{x^2-c^2}.\sqrt{x^2-a^2}$ $\implies (xb+ac)^2=(x^2-c^2)(x^2-a^2)$ $\implies x^2b^2+2xabc+a^2c^2=x^4-x^2(a^2+c^2)+a^2c^2 $ $\implies x^2b^2+2xabc=x^4-x^2(a^2+c^2)$ $\implies xb^2+2abc=x^3-x(a^2+c^2)$ $\implies x^3-x(a^2+b^2+c^2)-2abc=0$