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 February 7th, 2018, 08:50 PM #1 Banned Camp   Joined: Apr 2017 From: durban Posts: 22 Thanks: 0 Math Focus: Algebra Simultaneous equations i have look into the solving of the equation just want to confirm if my calculation is right. Q: X + Y + Z = 1 X^2 + Y^2 + Z^2 = 35 X^3 + Y^3 + Z^3 = 97 FIND THE VALUE OF X, Y AND Z Last edited by greg1313; February 8th, 2018 at 03:47 PM.
 February 7th, 2018, 08:56 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,203 Thanks: 1157 just looking at the second equation I see that $35 = 25 + 9 + 1$ making that work for the first equation we'd get $1 = 5 + (-3) + (-1)$ now let's check via the third equation $(5)^3 + (-3)^3 + (-1)^3 = 125 - 27 - 1 = 97$ hurrah $x = 5, ~y=-3,~z=-1$ or since the system is completely symmetric any permutation of those equalities.
February 8th, 2018, 12:40 AM   #3
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Quote:
 Originally Posted by romsek just looking at the second equation I see that $35 = 25 + 9 + 1$ making that work for the first equation we'd get $1 = 5 + (-3) + (-1)$ now let's check via the third equation $(5)^3 + (-3)^3 + (-1)^3 = 125 - 27 - 1 = 97$ hurrah $x = 5, ~y=-3,~z=-1$ or since the system is completely symmetric any permutation of those equalities.
How are you sure you found all the solutions?

 February 8th, 2018, 01:53 PM #4 Global Moderator   Joined: May 2007 Posts: 6,642 Thanks: 625 Is there an underlying condition that only integer solutions be considered? If so, you are done.
 February 8th, 2018, 02:04 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 2,203 Thanks: 1157 this is the basic algebra forum people. They aren't looking for real solutions of a non-linear system of equations.
February 8th, 2018, 07:28 PM   #6
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Quote:
 Originally Posted by romsek just looking at the second equation I see that $35 = 25 + 9 + 1$ making that work for the first equation we'd get $1 = 5 + (-3) + (-1)$ now let's check via the third equation $(5)^3 + (-3)^3 + (-1)^3 = 125 - 27 - 1 = 97$ hurrah $x = 5, ~y=-3,~z=-1$ or since the system is completely symmetric any permutation of those equalities.
well i try solving using substitution metrhod starting with equation 1 i.e
x = 1 -y - z------- eqn.1
substitute eqn. 1 for eqn. 2
x^2 + Y^2 + z^2 = 35
(1 - y - z)^2 + Y^2 + z^2 = 35

 February 8th, 2018, 11:54 PM #7 Global Moderator   Joined: Dec 2006 Posts: 19,983 Thanks: 1853 (1 - Y - Z)² + Y² + z² = 35 implies (Y + Z - 2)² + 2(Y + Z - 2) - 16 = (Y - 1)(Z - 1). From X³ + Y³ + Z³ = 97, one gets (1 - Y - Z)³ + Y³ + Z³ = 97, so 3Y² + 3Z² - 3Y²Z - 3YZ² + 6YZ - 3Y - 3Z = 96, which implies (Y - 1)(Z - 1)(Y + Z) = -32. Hence ((Y + Z - 2)² + 2(Y + Z - 2) - 16)(Y + Z) = -32, which implies Y³ + Z³ + 3Y²Z + 3YZ² - 2Y² - 2Z² - 4YZ - 16Y - 16Z + 32 = 0, which factorizes as (Y + Z + 4)(Y + Z - 4)(Y + Z - 2) = 0. Hence Y + Z = ±4 or 2. If Y + Z = 2, X = -1 and (Y - 1)(Z - 1) = -16. From there, it's easy to show that (Y, Z) = (-3, 5) or (5, -3). I'll leave you to finish from there.

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