It is known that the square of a + the square of B + the square of C = (a + B + C), ABC is not equal to 0, and the proof is: 1 / A + 1 / B + 1 / C = 0 /It's a division sign

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca
Because a & sup2; + B & sup2; + C & sup2; = (a + B + C) & sup2;
So 2Ab + 2BC + 2ca = 0
ABC is not equal to 0
Divide both sides by 2abc
1/a+1/b+1/c=0

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