# Find the numbers satisfying $x+y=19$ and $x^3+y^3=2071$

The sum of the cubes of two numbers is $2071$, while the sum of the two numbers themselves is $19$. Find the two numbers.

I've been working hard to solve this problem and I need someone to tell me how to solve it, it perplexes me.

I know the answer is $7$ and $12$ and it might have something to do with $$a^3 + b^3 = (a + b)(a^2 – ab + b^2)$$ but how do they get to that answer?

Any help will be appreciated thank you.

I followed this Yahoo question up to where it said solve it then I searched like $4$ hours looking for a way to solve it, so I'm hoping someone can help me out.

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So you have $$x+y=19$$ and $$x^2-xy+y^2=109$$ now just substitute $y = 19-x$ into the second equation, to get $$x^2-x(19-x)+(19-x)^2=109$$ After simplifying, we obtain $$x^2 - 19x+ 84=0$$

Solving this equation, we will get $x=7$ or $12$. Now, by symmetry, the $2$ numbers will be $7$ and $12$.

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Thanks! You made it really simple for me to follow. – user123940 Dec 9 '13 at 2:25

Recall the following identity: $$x^3+y^3 = (x+y)^3 - 3xy(x+y)$$ Hence, we get that $$2071 = 19^3 - 57xy \implies xy = 84$$ It is now slightly easier to solve $$x+y=19 \text{ and } xy=84$$ I trust you can finish it off from here. (Hint: Express $y$ as $\dfrac{84}x$ and solve the quadratic $x+\dfrac{84}x = 19$)

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Hint: Complete the square.

$$x^2-xy+y^2 = (x+y)^2 -3xy$$

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