1.

1/3abc from -2/3abc​

Answer»

Step-by-step explanation:

7a+B+c=1……………….(1)

ab+bc+ca=1/3…………(2)

ab+c(b+a)=1/3

ab+c(1-c)=1/3

ab+c-c^2=1/3…………….(3)

From eq.(2)

b(a+c)+ca=1/3

b(1-b)+ca=1/3

ca+b-b^2=1/3………………..(4)

Again from eq.(2)

bc+ca+ab=1/3

bc+a(c+b)=1/3

bc+a(1-a)=1/3

bc+a-a^2=1/3………………….(5)

subtract eq.(4)from(3)

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

a(b-c)-1(b-c)+(b-c)(b+c)=0

(b-c)(a-1+b+c)=0

Either b-c=0 => b=c………………(6)

Suntract eq.(5)from(4).

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

c(a-b)-1(a-b)+(a-b)(a+b)=0

(a-b)(c-1+a+b)=0

Either a-b=0 => a =b………………(7)

From eq.(7) and (6).

a=b=c =k(let)

a : b : c = k : k : k or 1 : 1 : 1 , answer.


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