triangle

In a triangle ABC, a,b,c are 3 opposite side of each angle A,B,C, if a^2-c^2=2b, and SinA*CosC=3*CosA*SinC, find b

$$\boxed{ (1)\quad a^2-c^2=2b\qquad (2) \quad SinA*CosC=3*CosA*SinC}$$

$$\small{\text{ sine rule: }} \frac{sinA}{a}=\frac{sinC}{c} \quad \Rightarrow \quad sinA=\frac{a}{c}\cdot sinC \\ (2):\qquad \frac{a}{c}\cdot sinC\cdot cosC = 3cosA\cdot sinC\Rightarrow \boxed{ a\cdot cosC=3c\cdot cosA }\\ \small{\text{ in every triangle: }}b=a\cdot cosC+c\cdot cosA\Rightarrow\boxed{ c\cdot cosA = b - a\cdot cosC }\\ a\cdot cosC=3 (b - a\cdot cosC)\\ 4\cdot a\cdot cosC = 3b\\ \boxed{ a\cdot cosC = \frac{3} {4}\cdot b }\\\\ \small{\text{ law of cosines: }} c^2=a^2+b^2-2ac\cdot cosC\\ (1): \qquad a^2-c^2=2ab\cdot cosC-b^2 =2b\\ 2ab\cdot cosC-b^2 =2b \quad | \quad :b \\ 2\cdot a\cdot cosC-b =2 \quad | \quad a\cdot cosC = \frac{3}{4}\cdot b \\ 2\cdot \frac{3}{4}\cdot b - b =2\\ \frac{3}{2}\cdot b - b =2 \quad | \quad \cdot 2 \\ 3b-2b=4\\ b=4$$

Brilliant as usual, heureka....!!!!