The value of an expression

Algebra Level 3

If for real numbers x , z , y x,z,y , there holds x z y = 1 xzy=1 , then the value of ( x + 1 x ) 2 + ( y + 1 y ) 2 + ( z + 1 z ) 2 − ( x + 1 x ) ( y + 1 y ) ( z + 1 z ) \Big(x+\dfrac1x\Big)^2+\Big(y+\dfrac1y\Big)^2+\Big(z+\dfrac1z\Big)^2-\Big(x+\dfrac1x\Big)\Big(y+\dfrac1y\Big)\Big(z+\dfrac1z\Big) is


The answer is 4.

A Former Brilliant Member by A Former Brilliant Member

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

Because of x y z = 1 xyz=1 , we have ( x + 1 x ) 2 + ( y + 1 y ) 2 + ( z + 1 z ) 2 − ( x + 1 x ) ( y + 1 y ) ( z + 1 z ) = x 2 + y 2 + z 2 + 1 x 2 + 1 y 2 + 1 z 2 + 6 − ( x 2 + 1 ) ( y 2 + 1 ) ( z 2 + 1 ) \Big(x+\dfrac1x\Big)^2+\Big(y+\dfrac1y\Big)^2+\Big(z+\dfrac1z\Big)^2-\Big(x+\dfrac1x\Big)\Big(y+\dfrac1y\Big)\Big(z+\dfrac1z\Big)=x^2+y^2+z^2+\dfrac1{x^2}+\dfrac1{y^2}+\dfrac1{z^2}+6-(x^2+1)(y^2+1)(z^2+1) , whence there also follows x 2 + y 2 + z 2 + x 2 y 2 + y 2 z 2 + z 2 x 2 + 6 − x 2 y 2 z 2 − 1 − x 2 − y 2 − z 2 − x 2 y 2 − y 2 z 2 − z 2 x 2 = 4 x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2+6-x^2y^2z^2-1-x^2-y^2-z^2-x^2y^2-y^2z^2-z^2x^2=4 .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...