I'm assuming you mean x^3 + y^3 = z^3, but if you don't, well suck it up. THere actually are two solutions, but both involve the nuber zero. If you are looking for a solution to this without a zero and only integers, it's impossible. There's a long and complicated proof for that but my brother took number theory, i havent yet. i just know that this isn't possible with integers and no zeros. The two cases where it does work is 0^3 +0^3 =0^3 (duh) and 1^3 +0^3 = 1^3
if you need more help, ill probably never see this again, so get owned
One equation with two unknowns usually does not have a solution.
It is an algebraic expression.
x3-y3
x^2 - xy + y^2
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
One equation with two unknowns usually does not have a solution.
3xyz
It is an algebraic expression.
x3-y3
(x2 - xy + y2)(x + y)
x^2 - xy + y^2
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
(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.
x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)
The required result will be 3xyz
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