Question

if a²=by +cz,b²=cz+ax,c²=ax+by,then the value of (x/a+x)+(y/b+y)+(z/c+z)​

Answer 1

Answer:

If x^2/by+cz = y^2/cz+ax = z^2/ax+by = 1, what is the value of a/a+x + b/b+y + c/c+z?

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x2by+cz=1=>x2=by+cz—(1)

y2cz+ax=1=>y2=cz+ax—(2)

z2ax+by=1=>z2=ax+by—(3)

(1)+(2)+(3):

x2+y2+z2=(by+cz)+(cz+ax)+(ax+by)=2ax+2by+2cz

=>x2−2ax+y2−2by+z2−2cz=0

=>(x−a)2+(y−b)2+(z−c)2−a2−b2−c2=0

=>(x−a)2+(y−b)2+(z−c)2=a2+b2+c2—(4)

Equation (4) is true when x=2a,y=2b and z=2c

aa+x=aa+2a=13

bb+y=bb+2b=13

cc+z=cc+2c=13

=>aa+x+bb+y+cc+z=13+13+13=1

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