Answer to Question #128177 in Algebra for Thembelani

Question

X²-y²-c²+2yc÷((x+y-c)/(x+y+c))

Answer 1

$simplify:\\ x^2-y^2-c^2+\frac{2yc}{\frac{x+y-c}{x+y+c}}=\\ =x^2-y^2-c^2+2yc\cdot \frac{x+y+c}{x+y-c}=\\ =x^2-y^2-c^2+2yc\cdot \frac{x+y-c+2c}{x+y-c}=\\ =x^2-y^2-c^2+2yc+\frac{4yc^2}{x+y-c}=\\ =x^2-(y-c)^2+\frac{4yc^2}{x+y-c}=\\ =(x-y+c)(x+y-c)+\frac{4yc^2}{x+y-c};$