How to solve the following pair of non-linear equations $x^3+y^3=152\\x^2y+xy^2=120$
Best regards !
 A: Multiply by three the second equation and add this to the first one, and you'll get (check this!)
$$(x+y)^3=152+360=512=2^9\implies x+y=2^3=8$$
Now take it from here.
A: Here is another way through, which simply depends on seeing the factor $x+y$ in both parts of what is given, reducing the cubic to a quadratic as follows (though the elimination of the constant term will always give a homogeneous cubic even without noticing this factor):
$x^3+y^3=(x^2-xy+y^2)(x+y)=152=8\times 19$
$x^2y+xy^2=(x+y)xy=120=8\times 15$
Multiply the first equation by $15$ and the second by $19$ to obtain:
$$15(x+y)(x^2-xy+y^2)=19xy(x+y)$$
Then we either have $x=-y$, which is easily eliminated, or divide through by $x+y\neq 0$ to give $$15x^2-34xy+15y^2=0$$ which factorises (quadratic formula) as $$(5x-3y)(3x-5y)=0$$
Either of these factors gives a linear relation which can be plugged into either of the original equations to give a solution.
A: Another method :
Suppose $x,y$ be the roots of  $z^2+az+b=0$
$x+y=-a$ and $xy=b$
$x^3+y^3=(x+y)^3-3xy(x+y)=-a^3+3ab=152$
$xy(x+y)=-ab=120$
$a^3=-152-3*120=-512$
$a=-8$  and  $b=120/8=15$
Solve of  $z^2-8z+15=0$  for $x$ and $y$
