Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the numbers.

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- Oct 11th 2008, 06:48 AMgreat_mathA good question
Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the numbers.

- Oct 11th 2008, 07:03 AMMoo
Hello,

$\displaystyle a^3+b^3=5$

$\displaystyle a^2+b^2=3$

Note that $\displaystyle a^3+b^3=(a+b)(a^2+b^2-ab)$

$\displaystyle \implies 5=(a+b)(3-ab)$

If a and b are integers, then there are a few possibilities :

3-ab=1 and a+b=5 (1)

3-ab=5 and a+b=1 (2)

3-ab=-1 and a+b=-5 (3)

3-ab=-5 and a+b=-1 (4)

From (1), we get ab=2, that is a=1 and b=2 for example. But a+b is not equal to 5. Impossible for integers.

From (2), we get ab=-2. We want a+b=1. --> a=-1 and b=2.

From (3), we get ab=4 --> a=-1 and b=-4

From (4), we get ab=8. Impossible for integers. - Oct 11th 2008, 07:36 AMJhevon
Alternatively, let the numbers be $\displaystyle x$ and $\displaystyle y$. we have

$\displaystyle x^3 + y^3 = 5$

$\displaystyle x^2 + y^2 = 3$

So as Moo said, $\displaystyle 5 = (x + y)(3 - xy)$

that is, $\displaystyle \boxed{5 = 3(x + y) - (x^2y + xy^2)}$

Now, $\displaystyle (x + y)^3 = x^3 + y^3 + 3(x^2y + y^2x)$

which means $\displaystyle \boxed{5 = (x + y)^3 - 3(x^2y + xy^2)}$

Now, let $\displaystyle a = x+ y$ and $\displaystyle b = x^2y + xy^2$, then we have the system

$\displaystyle 3a - b = 5$ .....................(1)

$\displaystyle a^3 - 3b = 5$ ....................(2)

Solve this system for $\displaystyle a$ and you have your answer (since $\displaystyle a$ is the sum of the two numbers)

I like your solution, Moo :D

But, of course, who said they had to be integers? - Oct 11th 2008, 07:40 AMMoo
- Oct 15th 2008, 08:57 AMsharon333
Thanks for the formulas.