10xy3 * 8x5y3 = (10*8)*(x*x5)*(y3*y3) = 80x6y6
4
No
n = x2 + 1 n = y3 - 1 x2 + 1 = y3 - 1 x2 = y3 - 2 x = (y3 - 2).5 This means that you need to find all cubes where subtracting 2 yields a square: Cubes: 1, 8, 27, 64, 125,... (-2): -1, 6, 25, 62, 123,... (27,25) works. n = 26
I am not sure, but I usually play games on y8.
6y-y3 = 3
-y3 + 7y3
x3-y3
72
27-y3 factored completely = 24
X2+Y3+15 All you can do is simplify it.
No
y times y times y (or y3)
30
12x2 - 24x + 9y
The GCF of y3, y7, y8 is y3
y3 x y3 - y (3)3 x 3(3) - 3 9 x 9 - 3 = ? 9 x 9= 81 81 - 3 = 78 I hope that solves your problem