Algebra #1

Algebra Level 3

a , b , c , x , y , z a,~b,~c,~x,~y,~z are complex numbers that satisfy below conditions.

a = b + c x − 2 , b = c + a y − 2 , c = a + b z − 2 a=\frac{b+c}{x-2},~b=\frac{c+a}{y-2},~c=\frac{a+b}{z-2} x y + y z + z x = 67 , x + y + z = 2017 xy+yz+zx=67,~x+y+z=2017

Find the value of x y z xyz .


The answer is -5913.

View wiki
Boi (보이) by Boi (보이) , 19, South Korea

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Boi (보이)
Jul 11, 2017

I'll make this solution as simple as possible.

x − 2 = b + c a , y − 2 = c + a b , z − 2 = a + b c x − 1 = a + b + c a , y − 1 = a + b + c b , z − 1 = a + b + c c 1 x − 1 = a a + b + c , 1 y − 1 = b a + b + c , 1 z − 1 = c a + b + c 1 x − 1 + 1 y − 1 + 1 z − 1 = 1 ( x − 1 ) ( y − 1 ) + ( y − 1 ) ( z − 1 ) + ( z − 1 ) ( x − 1 ) = ( x − 1 ) ( y − 1 ) ( z − 1 ) x y z − 2 ( x y + y z + z x ) + 3 ( x + y + z ) − 4 = 0 ∴ x y z = 2 ( x y + y z + z x ) − 3 ( x + y + z ) + 4 = − 5913 . \begin{aligned} &x-2=\frac{b+c}{a},~y-2=\frac{c+a}{b},~z-2=\frac{a+b}{c} \\ \\ &x-1=\frac{a+b+c}{a},~y-1=\frac{a+b+c}{b},~z-1=\frac{a+b+c}{c} \\ \\ &\frac{1}{x-1}=\frac{a}{a+b+c},~\frac{1}{y-1}=\frac{b}{a+b+c},~\frac{1}{z-1}=\frac{c}{a+b+c} \\ \\ &\frac{1}{x-1}+\frac{1}{y-1}+\frac{1}{z-1}=1 \\ \\ &(x-1)(y-1)+(y-1)(z-1)+(z-1)(x-1)=(x-1)(y-1)(z-1) \\ \\ &xyz-2(xy+yz+zx)+3(x+y+z)-4=0 \\ \\ &\therefore~xyz=2(xy+yz+zx)-3(x+y+z)+4=\boxed{-5913}. \end{aligned}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...