# linear-algebra multilinear-algebra

My knowledge of algebra are still insufficient to resolve this problem. Any help would be greatly appreciated solve the system of equations

$$xy(x+y)=30\\ x^3+y^3=35$$

help

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Your equation is symmetric in $x$ and $y$, so they can be written using $x + y$ and $xy$. If you can find out $x + y$ and $xy$, you will be able to find out $x$ and $y$. –  Secret Math Jun 16 '13 at 19:13
This is hardly multilinear-algebra. –  mrf Jun 16 '13 at 19:13

$(x+y)^3=x^3+y^3+3xy(x+y)$

$\Rightarrow (x+y)^3=35+90=125\Rightarrow (x+y)=5$

$\Rightarrow xy(x+y)=5xy=30\Rightarrow xy=6$

$(x-y)^2=(x+y)^2-4xy=25-24=1\Rightarrow (x-y)=\pm1$

case 1: $x-y=1$ and $x+y=5$ ,$\Rightarrow x=3,y=2$

case 2: $x-y=-1$ and $x+y=5$, $\Rightarrow x=2,y=3$

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is it clear @Rasa ???? –  Abhra Abir Kundu Jun 16 '13 at 19:18

Note that $$(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 = x^3+y^3 + 3xy(x+y) = 35 + 3 \cdot 30 = 125$$ and therefore $x+y = \sqrt[3]{125} = 5$ and $xy = \frac{30}{x+y} = 6$.

Thus $y = 5-x$ and $6 = xy = x(5-x)$, and so $x^2-5x+6=0$, which implies $x \in \{2,3\}$ and so $y \in \{3,2\}$. Thus, either $(x,y)=(2,3)$ or $(x,y) = (3,2)$...

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