Variables of multi-equations

May 2010
11
1
The question is:

\(\displaystyle a^3+6abd+3c(ac+b^2+d^2)=1\)
\(\displaystyle b^3+6abc+3d(a^2+bd+c^2)=0\)
\(\displaystyle c^3+6bcd+3a(ac+b^2+d^2)=0\)
\(\displaystyle d^3+6acd+3b(a^2+bd+c^2)=0\)

How to find the value of a,b,c,d?
Thanks
 

Soroban

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May 2006
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Hello, haedious!

\(\displaystyle \begin{array}{ccc}a^3+6abd+3c(ac+b^2+d^2) &=& 1 \\
b^3+6abc+3d(a^2+bd+c^2) &=& 0 \\
c^3+6bcd+3a(ac+b^2+d^2) &=& 0 \\
d^3+6acd+3b(a^2+bd+c^2) &=& 0 \end{array}\)

By inspection: .\(\displaystyle \begin{Bmatrix}a &=& 1 \\ b &=& 0 \\ c&=&0 \\ d&=&0 \end {Bmatrix}\)

 
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Reactions: haedious
May 2010
11
1
Thank you quickly reply, seem I mess up something...
But are there only the answer {1,0,0,0} to the question?
What if I knew the value of a is equal to 0?
 
Apr 2010
160
0
dfsdfdf
Dear Soroban,

Is there other ways of solving this question instead of by inspection.
I notice there is some form of symmetry in the given equations which is some kind of cyclic order.
Thanks
Kingman