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# Help!

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The sum of 3 real numbers is known to be zero. If the sum of their cubes is $$\pi^e$$, what is their product equal to?

Jun 9, 2019

#1
+28208
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As follows:

Jun 9, 2019
#2
+23357
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The sum of 3 real numbers is known to be zero.
If the sum of their cubes is $$\pi^e$$, what is their product equal to?

$$x+y+z = 0 \\ x^3+y^3+z^3 = \pi^e$$

1.)

$$\begin{array}{rcll} 0&=&(x+y+z)^3 \\ &=& x^3+y^3+z^3 +6xyz+3\left( x^2(y+z)+y^2(x+z)+z^2(x+y)\right) \end{array}$$

2.)

$$\begin{array}{rcll} 0&=&(x+y+z)(x^2+y^2+z^2) \\ &=& x^3+y^3+z^3 + x^2(y+z)+y^2(x+z)+z^2(x+y) \\ x^2(y+z)+y^2(x+z)+z^2(x+y) &=& -(x^3+y^3+z^3) \\ \end{array}$$

$$\begin{array}{rcll} x^3+y^3+z^3 +6xyz+3\left( x^2(y+z)+y^2(x+z)+z^2(x+y)\right) &=& 0 \\ x^3+y^3+z^3 +6xyz+3\left( -(x^3+y^3+z^3) \right) &=& 0 \\ -2(x^3+y^3+z^3) +6xyz &=& 0 \\ 6xyz &=& 2(x^3+y^3+z^3) \\ xyz &=& \dfrac{1}{3}(x^3+y^3+z^3) \\ \mathbf{xyz} &=& \mathbf {\dfrac{ \pi^e}{3}} \\ \end{array}$$

Jun 9, 2019