# Variables of multi-equations

#### haedious

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

MHF Hall of Honor
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: .$$\displaystyle \begin{Bmatrix}a &=& 1 \\ b &=& 0 \\ c&=&0 \\ d&=&0 \end {Bmatrix}$$

haedious

#### haedious

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?

#### kingman

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

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