# Tricky Problem

• Nov 6th 2009, 05:10 PM
acevipa
Tricky Problem
Ok so you are given; a, b, c are integers

abc = 729
and a + b + c = 91

Find a^2 + b^2 + c^2

How could I do this algebraically?
• Nov 6th 2009, 05:32 PM
Wilmer
Hint: 1+9+81=91
• Nov 6th 2009, 05:59 PM
Bacterius
Say $\displaystyle c = 1$. Therefore, we have :

$\displaystyle ab = 729$
$\displaystyle a + b = 90$

Substitute :

$\displaystyle a(90 - a) = 729$
$\displaystyle 90a - a^2 = 729$
$\displaystyle a^2 - 90a + 729 = 0$

Solve for $\displaystyle a$.

Facts for the lazy :
Spoiler:
You sure ?
Spoiler:

$\displaystyle \triangle = 90^2 - 2916 = 5184$

$\displaystyle a = \frac{\ {90 - \sqrt{\triangle}}}{2} = 9$

$\displaystyle b = \frac{\ {90 + \sqrt{\triangle}}}{2} = 81$

And for the curious :
Spoiler:
Since you would get a polynomial with the same coefficients if you had substituted $\displaystyle b$ instead of $\displaystyle a$, the two solutions of the quadratic equation are $\displaystyle a$ and $\displaystyle b$.
Oh, and by the way, you just factored 729 into smaller factors :p

Finding the sum of the squares shouldn't be too hard now ;) (remember $\displaystyle c = 1$)