1. ## Variables of multi-equations

The question is:

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

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

2. Hello, haedious!

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

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

3. 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?

4. 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