Question

if a^2+b^2+c^2=ab+bc+ca then find the value of (a+c)/b

Don't post irrelevant answer!!​

Answer 1

Step-by-step explanation:

2(a^2+b^2+c^2-ab-bc-ca)=0

2a^2+2b^2+2c^2-2ab-2bc-2ca=0

(a^2-2ab+b^2)+(b^2-2ab+c^2)+(c^2-2ca+a^2)=0

(a-b)^2+(b-c)^2+(c-a)^2=0

since the sum of two square is 0then each term should be 0.

(a-b)^2=0,(b-c)^2=0,(c-a)^2=0

(a-b)=0,(b-c)=0,(c-a)=0

a=b,b=c,c=a

a=b=c.

i.e c+a/b=2

Answer 2

Answer:

2

Step-by-step explanation:

here we have given,

${a}^{2} + {b}^{2} + {c }^{2} = ab + bc + ca$

therefore

${a}^{2} = ab \\ a = b$

and also

${b}^{2} = bc \\ b = c$

and

${c}^{2} = ca \\ c = a$

therefore we get from above equation,

a = b = c

therefore,

$(a + c) \div b$

$= (a + a) \div a$

$= 2a \div a$

$= 2$