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