x^2/y^3 = x^2*y^(-3)
(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)
208
the answers is d
8 x y1.5 sqrt(2)
X2+Y3+15 All you can do is simplify it.
x^2/y^3 = x^2*y^(-3)
x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)
(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)
208
(x2 - xy + y2)(x + y)
the answers is d
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
√(49x4y6) = √49√x4√y6 = 7*x2*y3
8 x y1.5 sqrt(2)
Not sure. There is no particular name for x2 and 2xy (both include x) in x2 + 2xy + y2 + y3
the equation is A= y2-y2/x3-x2 after that you find the y-intercept by doing, b= y1+y2+y3-A(x1+x2+x3)/3