Find all solutions of the system: (hint: factor the left
side of the second equation:
x-y = 3
x^3 - y^3 = 387
So now after factoring I have this new system:
x-y = 3
(x-y)(x² + xy + y²) = 387
but now how do I continue the problem?
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Since you have x-y = 3 you can substitute 3 for (x-y) in
(x-y)(x² + xy + y²) = 387
getting
3(x² + xy + y²) = 387
Divide both sides by 3
x² + xy + y² = 129
Now since x-y = 3, solving for y gives
y = x-3
So substitute (x-3) for y in
x² + xy + y² = 129
x² + x(x-3) + (x-3)² = 129
x² + x² - 3x + (x-3)(x-3) = 129
2x² - 3x + (x²-6x+9) = 129
2x² - 3x + x² - 6x + 9 = 129
3x² - 9x - 120 = 0
3(x² - 3x - 40) = 0
3(x - 8)(x + 5) = 0
That gives two possible values for x:
x = 8 and x = -5
Now substituting x = 8 into
y = x - 3
y = 8 - 3
y = 5
So one solution is (x, y) = (8, 5)
Substituting x = -5 into
y = x - 3
y = -5 - 3
y = -8
So the other solution is (x, y) = (-5, -8)
Edwin
