anemone said:
The real numbers $a,\,b,\,c$ and $d$ satisfy the following simultaneous equations:
$abc-d=1\\bcd-a=2\\cda-b=3\\dab-c=-6$
Prove that $a+b+c+d \ne 0$.
[sp]abc-d=1......(1)
bcd-a=2......(2)
cda-b=3.......(3)
dab-c=-6.......(4)
Supose a+b+c+d=0...(5)
then by adding all the above equations we get:
abc+bcd+cda+dab=0=> bc(a+d)+da(c+b)=0......(6)
But from(5) we have :a+d=-(c+b)............(7)
Substituting (7) into (6) we have : -bc(c+b)+da(c+b)=0=>bc(c+b)-da(c+b)=0=>(c+b)(bc-ad)=0.....(8)
This equality now must give us a contradiction
But(c+b)(bc-ad)=0.=> c=-b or bc=ad
Hence for c=-b the equality (3) becomes : -bda+c=3 and adding to that eqaulity (4) we have:-3=0 a contradiction
Now we have find out if bc=ad give us a contradiction
Now again from(5) we have : $(b+c)^2=(-(a+d))^2\Leftrightarrow b^2+c^2+2bc=a^2+d^2+2ad$...(9)
But since bc=ad (9) becomes :$(b^2-a^2)+(c^2-d^2)=0\Rightarrow (b+a)(b-a)+(c+d)(c-d)=0$.......(10)
But from (5) again we have: (b+a)=-(c+d)..............(11)
And substituting (11) into (10)we have :-(c+d)(b-a)+(c+d)(c-d)=0=>(c+d)(c-d)-(c+d)(b-a)=0=>(c+d)(c-d-(b-a))=0=>(c+d)(c-d-b+a)=(c+d)(c+a-(d+b))=(c+a)(-2(d+b))=0 since from (5) we have (c+a)=-(d+b)
Hence(c+a)(d+b)=0 => c=-a or d=-b
For c=-a (2) becomes -dab+c=2 and adding to it equality (6) wehave: -4=0 a contradiction
for d=-b (3) becomes -cba+d=2 and adding to it equality(1) we have : 4=0 a contradiction again
Hence a+b+c+d is different from zero[/sp]