For real numbers x and y, a new operation *, X * y = ax + bx-3, where a and B are constants, if 1 * 2 = 9, (- 3) * 3 = 6, find the value of 2 * (- 7)

For real numbers x and y, a new operation *, X * y = ax + bx-3, where a and B are constants, if 1 * 2 = 9, (- 3) * 3 = 6, find the value of 2 * (- 7)

I think the original problem may be x * y = ax + by-3! From the problem, we know that 1 * 2 = a + 2b-3 = 9; (- 3) * 3 = - 3A + 3b-3 = 6, the solution of a = 2; b = 5, the identity of the original equation becomes x * y = 2x + 5y-3, so 2 * (- 7) = 2 * 2 + 5 * (- 7) - 3 = - 34

（ax+by)²+（bx-ay)² （a+b)(a-b)+4(b-1)

（ax+by)²+（bx-ay)²
=a²x²+2abxy+b²y²+b²x²-2abxy+a²y²
=(a²x²+a²y²)+(b²x²+b²y²)
=a²(x²+y²)+b²(x²+y²)
=(a²+b²)(x²+y²)
（a+b)(a-b)+4(b-1)
=a²-b²+4b-4
=a²-(b²-4b+4)
=a²-(b-2)²
=(a-b+2)(a+b-2)

（x²+x)²-26(x²+x)+120 ax+ay+bx+by-a-b x²(x²-y²)-72y⁴

（x²+x)²-26(x²+x)+120
=(x²+x-6)(x²+x-20)
=(x+3)(x-2)(x+5)(x-4)
ax+ay+bx+by-a-b
=a(x+y)+b(x+y)-(a+b)
=(a+b)(x+y)-(a+b)
=(a+b)(x+y-1)
x²(x²-y²)-72y⁴
=x⁴-x²y²-72y⁴
=(x²+8y²)(x²-9y²)
=(x²+8y²)(x+3y)(x-3y)