(x^2+2x+ 1)+(y^2-6y+9)=0

(x+ 1)^2+(y-3)^2=0

The square is greater than or equal to 0.

The total is 0.

So both squares are equal to 0.

So x+ 1 = 0 and y-3 = 0.

x=- 1，y=3

xy=-3

(ax+2)(x^2+ 12x- 1 1)

Item X 2 is

ax* 12x+2*x^2=( 12a+2)x^2

So 12a+2=0.

a=- 1/6

(a-2)^2(a^2-6a-9)-a(a^2-2a- 15)

=(- 1/6-2)^2*( 1/36+ 1-9)-(- 1/6)*(a-5)(a+3)

= 169/36*(-287/36)+ 1/6*(- 1/6-5)*(- 1/6+3)

=-5 1665/ 1296

Did you copy this question wrong?