Let's Solve A Cool Cubic System

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[FisicTrapella]

2 nd method
If you subtract instead of adding you'll get (x-y)^3 = 1. So x-y = 1.
x+y = 3
x-y = 1
That gives x=2, y=1

[mcwulf]

Add 'em, subtract 'em.
Perfect cubes.
y = cube roots of unity
x = 2y.

[SidneiMV]

extremely simple and easy

x³ + 3xy² = 14 (I)
y³ + 3x²y = 13 (II)

(I) + (II)
x³ + 3xy(x + y) + y³ = 27
(x + y)³ = 27 => x + y = 3 (III)

(I) - (II)
x³ - 3x²y + 3xy² - y³ = 1
(x - y)³ = 1 => x - y = 1 (IV)

(III) + (IV)
x + y = 3
x - y = 1
2x = 4 => *x = 2*
x + y = 3 => y = 3 - x => *y = 1*

[martinmohle]

In 2nd method you could as a 2nd step subtract 2nd from 1st eqtn wich leads to (x-y)³=1 => system of linear eqns
I) x+y=3
II) x-y=1