1.75
0.5
y times y times y
If x = 3 and y = 4 then the answer is 2
Yes. You need at least three points with x and y coordinates for this. Let the points be: A(x1, y1), B(x2, y2) and C(x3, y3). For each of the points yi = ax2i + bxi + c, so you have to solve a set of simultaneous equations for i=1 ... 3(the y's and x's are no longer variables! the a, b, c are!): y1 = ax21 + bx1 + c, y2 = ax22 + bx2 + c, y3 = ax23 + bx3 + c
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)
(x - y)(x^2 + xy + y^2
(x + y)3 + (x - y)3 = (x3 + 3x2y + 3xy2 + y3) + (x3 - 3x2y + 3xy2 - y3) = 2x3 + 6xy2 = 2x*(x2 + 3y2)
When it is of the form x3 + y3 or x3 - y3. x or y can have coefficients that are perfect cubes, or even ratios of perfect cubes eg x3 + (8/27)y3.
0
(x2 - xy + y2)(x + y)
x^2 - xy + y^2
y^3
( 6x2 / y ) × ( y3 / 12x4) = 6x2y3 / 12x4y = y2 / 2x2
You would need to know the value of either x or y, so that you could solve for the one independent variable.
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