Re: [中學] 對稱多項式
※ 引述《Ericdion ( 由心出發 )》之銘言：
: 兩題，第296與309
: 想了幾天還是沒找到方法
: 請高手指教，感恩！！！
: http://i.imgur.com/1tTQBn5.jpg
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: Sent from JPTT on my Asus PadFone T004.
309.
(x + y + z) * (x^2 + y^2 + z^2)
= (x^3 + y^3 + z^3) + [x^2(y + z) + y^2(z + x) + z^2(x + y)]
= (x^3 + y^3 + z^3) - 24
(x + y + z)^2
= (x^2 + y^2 + z^2) + 2(xy + yz + zx)
=> 9 = (x^2 + y^2 + z^2) + 2(xy + yz + zx)
1/x + 1/y + 1/z = (xy + yz + zx)/(xyz) = -1/3
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
= (x^3 + y^3 + z^3) - 24 + (x + y + z)*(- xy - yz - zx)
=> 9(xy + yz + zx) = - 24 + (-3)*(- xy - yz - zx)
=> xy + yz + zx = -4
=> 9 = (x^2 + y^2 + z^2) + 2(xy + yz + zx)
= (x^2 + y^2 + z^2) - 8
=> x^2 + y^2 + z^2 = 17
=> (x^3 + y^3 + z^3) - 24 = (x + y + z) * (x^2 + y^2 + z^2)
= -3 * 17
= -51
=> x^3 + y^3 + z^3 = -27
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03/18 21:59, , 1F
03/18 21:59, 1F
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