2023 SSMO Accuracy Round Problems/Problem 3

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. (Error compiling LaTeX. Unknown error_msg)

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

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2023 SSMO Accuracy Round Problems/Problem 3

Problem

Solution