Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

enter image description here

share|cite|improve this question
up vote 7 down vote accepted

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 $12$.

share|cite|improve this answer
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 $x+\dfrac{84}x = 19$)

share|cite|improve this answer

Hint: Complete the square.

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.