Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 3"

Latest revision as of 14:51, 9 September 2025

Problem

Suppose that $a, b, c$ are real numbers such $\begin{align*} a + b - c &= 4 \\ a^2 + b^2 + c^2 &= 14 \\ a^3 + b^3 - c^3 &= 34 \\ \end{align*}$ Find the sum of all possible values of $a+b+c$ .

Solution

Let $x = -c.$ Therefore, we have

$\begin{align} a+b+x&=4,\\ a^2+b^2+x^2&=14,\text{ and }\\ a^3+b^3+x^3 &= 34. \end{align}$

Subtracting equation (2) from the square of (1), we have $2(ab+ax+bx) = 2\implies ab+ax+bx = 1.$ Multiplying this by the first equation, we have $(a^2b+ab^2+ax^2+a^2x+b^2x+bx^2)+3abx = 4.$ Cubing the first equation, we have

\begin{align*} (a^3+b^3+x^3)+3(a^2b+ab^2+ax^2+a^2x+b^2x+bx^2)+6abx &= \\ (a+b+x)^3\implies34+3(4-3abx)+6abx = 64\implies abx=6. \end{align*}

Thus, $a,b,x$ are the roots of the polynomial $P(y) = y^3-4y^2+y-6,$ which factors $(y-3)(y-2)(y+1).$ So, the possible values of $a+b+c$ are

$\begin{align*} a+b+c = a+b-x = 3+2-(-1) = 6,\\ a+b+c = a+b-x = 3+(-1)-(2) = 0,\text{ and }\\ a+b+c = a+b-x = 2+(-1)-(3) = -2. \end{align*}$

Thus, the answer is $6+0+(-2) = \boxed{4}.$

~SMO_Team

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Difference between revisions of "2023 SSMO Accuracy Round Problems/Problem 3"

Latest revision as of 14:51, 9 September 2025

Problem

Solution

@@ Line 34: / Line 34: @@
 a+b+c = a+b-x = 3+2-(-1) = 6,\\
 a+b+c = a+b-x = 3+(-1)-(2) = 0,\text{ and }\\
-a+b+c = a+b-x = 2+(-1)-(3) = -2.</cmath>
+a+b+c = a+b-x = 2+(-1)-(3) = -2.
+\end{align*}</cmath>
 Thus, the answer is <math>6+0+(-2) = \boxed{4}.</math>
 ~SMO_Team