Let x , y , z be complex numbers such that the following equations hold. ⎩ ⎪ ⎨ ⎪ ⎧ x + y + z x 2 + y 2 + z 2 x y z = 2 = 3 = 4 Find the value of x y + z − 1 1 + y z + x − 1 1 + z x + y − 1 1 .
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Very nice solution and very intuitive
Nice solution ⌣ ¨
Nice solution :-)
Nice solution
x y = z 4
Hence first term becomes :
z 4 + z − 1 1 = z 2 − z + 4 z
Hence our desired sum becomes :
S = c y c ∑ z 2 − z + 4 z
Now 2 ( x y + y z + x z ) = ( x + y + z ) 2 − ( x 2 + y 2 + z 2 )
⇒ x y + y z + x z = 2 1
We assume x,y,z to be roots of a cubic polynomial hence the polynomial is given by :
p 3 − ( c y c ∑ x ) p 2 + ( c y c ∑ x y ) p − x y z = 0
Putting the given values we get :
2 p 3 − 4 p 2 + p − 8 = 0
Since z is a root hence we have :
2 z 3 − 4 z 2 + z − 8 = 0
Factorising we get :
2 ( z − 1 ) ( z 2 − z + 4 ) = 9 z
Hence we have :
9 2 ( z − 1 ) = z 2 − z + 4 z
Our desired sum becomes :
S = 9 2 c y c ∑ ( z − 1 )
Which it turns out to be :
S = 9 2 ( x + y + z − 3 ) = 9 − 2
Can you pls explain how you factorized 2 z 3 − 4 z 2 + z − 8 = 0 to 2 ( z − 1 ) ( z 2 − z + 4 ) = 9 z , I mean how did you know that you had to isolate 9z and keep it on RHS .
otherwise I liked your solution .
Pls help me : @SanjeetRaria ,@CalvinLin,@SandeepBharadwaj,@PiHanGoh,@MeghChoksi,@ArronKau,@MichaelMendrin,@Chew-SeongCheong .
This is a good question .
i did the exact same!!!!
Nice solution.
Thanks for appreciating the solution.
Nice way to use polynomial... I solved it similar to chew's. But i like your solution.
Nice solution Ronak! ⌣ ¨
can u please explain transition from : eq--> p^3-sigma(x)p^2+sigma(xy)p-xyz=0 TO 2p^-4p^2+p-8
Thats way more innovative, i expressed denominators as (x-1)(y-1) and so on then took LCM and added and inputted values after grouping terms accordingly,,,
This is my problem, Complicated Complex Fractions !!!
Welllll I think it'd be a better move to remove that link or your problem's rating would go down since anyone who clicked 'View Solution' here can get your problem right. @Sharky Kesa
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Yeah, I find it a little while ago. I'm sorry for posting it again. But I was not aware of this problem has been posted on Brilliant. @Sharky Kesa
This answer is checked to have no cyclic order differences.
x = -0.229220113606684-1.25479322434906i
y = -0.229220113606684+1.25479322434906i
z = 2.45844022721337
2
3.00000000000003
3.99999999999999
All checked correct.
0.324097828269638
-0.27316002524593-0.278842938744237i
-0.27316002524593+0.278842938744237i
Sum = -0.222222222222222 such that -2/ 9
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Since x + y + z = 2 , then z = 2 − x − y .
⇒ x y + z − 1 = x y + 2 − x − y − 1 = x y − x − y + 1 = ( x − 1 ) ( y − 1 )
⇒ x y + z − 1 1 + y z + x − 1 1 + z x + y − 1 1
= ( x − 1 ) ( y − 1 ) 1 + ( y − 1 ) ( z − 1 ) 1 + ( z − 1 ) ( x − 1 ) 1
= ( x − 1 ) ( y − 1 ) ( z − 1 ) x − 1 + y − 1 + z − 1 = ( x − 1 ) ( y − 1 ) ( z − 1 ) − 1
Now ( x − 1 ) ( y − 1 ) ( z − 1 ) = x y z − ( x y + y z + z x ) + ( x + y + z ) − 1 .
And x y + y z + z x = 2 1 [ ( x + y + z ) 2 − ( x 2 + y 2 + z 2 ) ] = 2 1 ( 4 − 3 ) = 2 1
Therefore,
x y + z − 1 1 + y z + x − 1 1 + z x + y − 1 1
= x y z − ( x y + y z + z x ) + ( x + y + z ) − 1 − 1
= 4 − 2 1 + 2 − 1 − 1 = 2 9 − 1 = − 9 2 ≈ − 0 . 2 2 2