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MIE 2016/Day 1/Problem 9

Problem 9

Let $x$, $y$ and $z$ be complex numbers that satisfies the following system:

$\begin{cases}x+y+z=7\\x^2+y^2+z^2=25\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{4}\end{cases}$


Compute $x^3+y^3+z^3$.


(a) $210$

(b) $235$

(c) $250$

(d) $320$

(e) $325$


Solution

We start by expanding $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$.


As we are given $(x+y+z)$ and $x^2+y^2+z^2$, we get $(xy+yz+xz)$ is $12$.


Next, we simplify the third case and obtain $4(xy+yz+xz)=xyz$


As we know $(xy+yz+xz)$ is $12$, we know $xyz$ is $48$


Next we expand $(x+y+z)^3=x^3+y^3+z^3+3((x+y+z)(xy+yz+xz)-xyz)$


Rearranging the equation we arrive at $x^3+y^3+z^3=(x+y+z)^3-3((x+y+z)(xy+yz+xz)-xyz)$


Substituting all the known values, we get $x^3+y^3+z^3=343-3((7)(12)-48) = 235$. $\boxed{\textbf{B}}$.

See Also

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