\(a,P=\left(3x^2+6xy+3y^2\right)-2\left(x+y\right)-100\\ P=3\left[\left(x+y\right)^2-2\cdot\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}\right]-\dfrac{301}{3}\\ P=3\left(x+y-\dfrac{1}{3}\right)^2-\dfrac{301}{3}\\ P=3\left(5-\dfrac{1}{3}\right)^2-\dfrac{301}{3}=3\cdot\left(\dfrac{14}{3}\right)^2-\dfrac{301}{3}=\dfrac{196}{3}-\dfrac{301}{3}=-35\)
\(b,A=\left(x+y\right)^2-2xy=10^2-2\cdot50=0\)