It is x^2 + xy + y^2
(x + y)(x + y)(x + y)
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
You have to put your heart into it!
find x3+y2 if x=4 and y=3 ( 4*4*4)+(3*3) 64 +9 =73
A cubic function.
It is x^2 + xy + y^2
(x + y)(x + y)(x + y)
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
x times x times x equals x cubed
You have to put your heart into it!
find x3+y2 if x=4 and y=3 ( 4*4*4)+(3*3) 64 +9 =73
64 x cubed minus y cubed is (4x - y)(16x^2 + 4xy + y^2)
There are an infinite number of real solutions. For example, x = 1, y = 1 and z = cuberoot(2), or x = 1, y = 2 and z = cuberoot(9). There are no integer solutions, as proven by Fermat's Last Theorem.
x=1
X^4
Divide both sides of the equation by Y2.That will leave you with X / Y2 = Y* * * * *That doesn't quite do it, does it? Try this:Take the cube root of both sides.y = ³√x.