Raymond B. answered 07/06/21
Math, microeconomics or criminal justice
12(x+y)^3/9(x+y) maybe 12=p, 9=q 3=c? x=a, b=y p(a+b)^c/q(a+b)
=[p(a+b)^c]/q(a+b)
or cancel the common factor (x+y)
= 12(x+y)^2/9 and divide 3 into 12 and 9 to get
=(4/3)(x+y)^2 = (p/q)(a+b)^c if p=4, q=3, c=2, a=x, b=y
=(4/3)(x^2 +2xy + y^2)
=(p/q)(a^c + cab + b^c)
Just a couple guesses. one might be what you want