Question

(x-b-c)/a + (x-c-a)/b + (x-a-c)/c = 3
Prove this, how becoming 3

Answer 1

$\frac{(x-b-c)}{a}+\frac{(x-a-b)}{c}+\frac{(x-c-a)}{b}=3\\\\bc(x-b-c)+ab(x-a-b)+ac(x-c-a)=3abc\\\\bcx-b²c-bc² +abx-a²b -ab² +acx -a²c -ac²=3 abc\\\\abx+bcx +cax=3abc+ a²b +ab²+ b²c+ bc²+ a²c+ ac²\\\\x(ab+bc+ac)= abc+a²b +ab²+ abc+ b²c +bc²+ abc +a²c +ac²\\\\x(ab+bc+ac)=ab(c+a+b)+bc(a+b+c)+ac(b+a+c)\\\\x(ab+bc+ac) =(a+b+c) (ab +bc+ ac)\\\\x =\frac{(a+b+c)(ab+bc+ac)}{(ab+bc+ac)}\\\\x=a+b+c$

PUTTING THE VALUE OF X IN THE EQUATION

$\frac{(a+b+c-b-c)}{a}+\frac{(a+b+c-a-b)}{c}+\frac{(a+b+c-c-a)}{b}=3\\\\\frac{a}{a}+\frac{c}{c}+\frac{b}{b}=3\\\\1+1+1=3\\\\3 = 3$

THIS IS YOUR REQUIRED ANSWER ☺️✌️