A good question

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October 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.
October 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.

$a^3+b^3=5$
$a^2+b^2=3$

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

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

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.
October 11th 2008, 07:36 AM
Jhevon

Alternatively, let the numbers be $x$ and $y$ . we have

$x^3 + y^3 = 5$
$x^2 + y^2 = 3$

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

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

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

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

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

$3a - b = 5$ .....................(1)
$a^3 - 3b = 5$ ....................(2)

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

I like your solution, Moo :D

But, of course, who said they had to be integers?
October 11th 2008, 07:40 AM
Moo

Quote:

Originally Posted by Jhevon

Alternatively, let the numbers be $x$ and $y$ . we have

$x^3 + y^3 = 5$
$x^2 + y^2 = 3$

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

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

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

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

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

$3a - b = 5$ .....................(1)
$a^3 - 3b = 5$ ................(2)

Solve this system for $a$ and you have your answer (since $a$ 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)
October 15th 2008, 08:57 AM
sharon333

Thanks for the formulas.