A Nice Algebra Problem | How to solve for a , b and c

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[@zonked1200]

You didn't verify that the solution is also -square root of 2, but it does work out. -root(2) + -root(2) = -2*root(2) and -root(2) * -root(2) * -root(2) comes out to be -2*root(2).

[@ครัวหนุ่มพม่า]

for all = 0

[@tomtke7351]

a+b
=
b+c
=
c+a
2a+2b+2c=3abc
a+b+c = (3/2)abc
Deadend
Restart
a+b = b+c
a = c eq.4
b+c = c+a
b = a eq.5
c+a = a+b
c = b eq.6
Or
a = b = c eq.7
Simplify
a+b = abc
a+a = a^3
2a= a^3
2 = a^2
a = +/-sqrt(2) eq.8
Verify
sqrt(2)+sqrt(2)=?(sqrt(2))^3
2sqrt(2)=♥️2sqrt(2)✔️
AND
(-sqrt(2))+(-sqrt(2))
=? (-sqrt(2))^3
-2sqrt(2)=2(-sqrt(2))
-2sqrt(2)=♥️-2sqrt(2)✔️
FINAL ANSWER:
a = b = c = +/- sqrt(2)

[@mathematicsformankind-onlinecl]

Take a=0, b=√2 then is it possible 0+√2=0??

[@researchinbreeder]

No. Each possible answer is mutually exclusive. "a" (and therefore "b" and "c") can be 0, OR root 2, OR negative root 2. All three answers solve the given equations.
0+0 = 0^3
r2+r2 = r2^3
-r2+-r2 = -r2^3

[@henkn2]

No indeed. a=b=c