Three equations, three variables, three powers
⎩ ⎪ ⎨ ⎪ ⎧ x + y + z = 1 0 x 2 + y 2 + z 2 = 2 0 x 3 + y 3 + z 3 = 1 0 0
Consider three complex numbers x , y and z that satisfy the system of equations above. Find the value of x y z .
The answer is 100.
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3 solutions
That's the proper and standard way to solve these type of questions. !!😀
We can use Girard's formula, x y z = 6 1 ( p 1 3 − 3 p 1 p 2 + 2 p 3 ) = 1 0 0 , where p k = x k + y k + z k
Sounds interesting. Where can I read more about Girard's formula? (Tried googling it but found no obvious links)
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These are usually called " Newton Sums " or "Newton Identities" in the Anglo-Saxon world, but the identity we are using here was know to Albert Girard long before Newton was born ;)
Squaring the first equation and subtracting second from it we get (xy+yz+zx)=40.
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Substituting the values from three equations and the one we found,
100 - 3xyz=10*(20-40), we get xyz=100.
→ x + y + z = 1 0 .... ( 1 )
→ x 2 + y 2 + z 2 = 2 0 .... ( 2 )
→ x 3 + y 3 + z 3 = 1 0 0 .... ( 3 )
Squaring both sides to ( 1 ) .
x 2 + y 2 + z 2 + 2 ( x y + y z + z x ) = 1 0 0
2 0 + 2 ( x y + y z + z x ) = 1 0 0
x y + y z + z x = 4 0
Adding ( − 3 x y z ) both sides to ( 3 ) .
x 3 + y 3 + z 3 − 3 x y z = 1 0 0 − 3 x y z
( x + y + z ) ( x 2 + y 2 + z 2 − ( x y + y z + z x ) ) = 1 0 0 − 3 x y z
1 0 × ( 2 0 − 4 0 ) = 1 0 0 − 3 x y z
x y z = − 3 − 3 0 0 = 1 0 0