# A good question

• Oct 11th 2008, 06:48 AM
great_math
A 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 AM
Moo
Hello,
Quote:

Originally Posted by great_math
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.

\$\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 AM
Jhevon
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 AM
Moo
Quote:

Originally Posted by Jhevon
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?

Yours is far better... I wasn't able to finish off (Speechless)
• Oct 15th 2008, 08:57 AM
sharon333
Thanks for the formulas.