Hailie Blevins

2022-06-25

How to solve the following pair of non-linear equations
${x}^{3}+{y}^{3}=152\phantom{\rule{0ex}{0ex}}{x}^{2}y+x{y}^{2}=120$

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America Barrera

Expert

Multiply by three the second equation and add this to the first one, and you'll get
$\left(x+y{\right)}^{3}=152+360=512={2}^{9}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x+y={2}^{3}=8$
Now take it from here.

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Lydia Carey

Expert

Another method:
Suppose $x,y$ be the roots of ${z}^{2}+az+b=0$
$x+y=-a$ and $xy=b$
${x}^{3}+{y}^{3}=\left(x+y{\right)}^{3}-3xy\left(x+y\right)=-{a}^{3}+3ab=152$
$xy\left(x+y\right)=-ab=120$
${a}^{3}=-152-3\ast 120=-512$
$b=120/8=15$
Solve of ${z}^{2}-8z+15=0$ for $x$ and $y$

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