We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
155
2
avatar+1206 

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
avatar+28208 
+3

As follows:

 

 Jun 9, 2019
 #2
avatar+23357 
+3

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.)

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

 

 

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

 

laugh

 Jun 9, 2019

30 Online Users