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Roots of the equation is given by -b +-√b²-4ac/2a
-(b-c)+√(b-c)²-4(c-a)(a-b)/2(a-b)=-(b-c)-√(b-c)²-4(c-a)(a-b)  both roots are equal
squaring both side 
b²+c²-2bc= 4ac-4bc-4a²-4ab
(b+c)²=4(ac-ab -a²)       ac=ab because both roots are equal
taking root on both side to make a²  +ve
sp √(b+c)²=√4a²
hence b+c=2a
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