Give : \(\dfrac{bz+cy}{x\left(-ax+by+cz\right)}=\dfrac{cx+az}{y\left(ax-by+cz\right)}=\dfrac{ay+bx}{z\left(ax+by-cz\right)}\)
Prove that : a) \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{ay+bx}{b}\)
b) \(\dfrac{x}{a\left(b^2+c^2-a^2\right)}=\dfrac{y}{b\left(a^2+c^2-b^2\right)}=\dfrac{z}{c\left(a^2+b^2-c^2\right)}\)
