Cubic Polynomial Solution 2
Let $ a, $ $ b, $ and $ c $ be real numbers such that $ ab + ac + bc = 0 $ and $ (a + b + c + 1)^2 = abc $. Find all possible values of
\[(ab - c)(ac - b)(bc - a).\]Enter the possible values, separated by commas. For example, if you think the possible values are 3, 4, and 5, enter "3, 4, 5", without the quotation marks.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
From $ ab + ac + bc = 0, $ we get $ ab = -ac - bc, $ $ ac = -ab - bc, $ and $ bc = -ab - ac $. Then
\begin{align*}
(ab - c)(ac - b)(bc - a) &= (-ac - bc - c)(-ab - bc - b)(-ab - ac - a) \\
&= -abc(a + b + 1)(a + c + 1)(b + c + 1).\end{align*}Let $ s = a + b + c $. Then
\[-abc(a + b + 1)(a + c + 1)(b + c + 1) = -abc(s + 1 - c)(s + 1 - b)(s + 1 - a).\]We know that $ a, $ $ b, $ and $ c $ are the roots of the polynomial
\[p(x) = (x - a)(x - b)(x - c).\]Expanding, we get
\[p(x) = x^3 - (a + b + c) x^2 + (ab + ac + bc)x - abc.\]We know that $ ab + ac + bc = 0 $. Also, $ abc = (a + b + c + 1)^2 = (s + 1)^2, $ so
\[p(x) = x^3 - sx^2 - (s + 1)^2.\]Setting $ x = s + 1, $ we get
\[p(s + 1) = (s + 1)^3 - s(s + 1)^2 - (s + 1)^2 = 0.\]But
\[p(s + 1) = (s + 1 - a)(s + 1 - b)(s + 1 - c).\]Therefore,
\[-abc(s + 1 - c)(s + 1 - b)(s + 1 - a) = 0.\]The only possible value of the given expression is $ \boxed{0} $. The triple $ (a,b,c) = (1,-2,-2) $ shows that the value of 0 is achievable.
Freeflows