標題:
Challenge!
發問:
Please help! 1. Solve x if (3a^2 + b^2) (x^2 - x + 1) = (3b^2 + a^2) (x^2 + x + 1). 2. If x^1/3 + y^1/3 + z^1/3 = 0, show that (x+y+z)^3 = 27xyz Thanks!
(1) If a≠b and a≠-b, (3a2 + b2) (x2 - x + 1) = (3b2 + a2) (x2 + x + 1) (3a2 + b2) x2 - (3a2 + b2) x + (3a2 + b2) = (3b2 + a2) x2 + (3b2 + a2) x + (3b2 + a2) (2a2 - 2b2) x2 - (4a2 + 4b2) x + (2a2 - 2b2) = 0 (a2 - b2) x2 - (2a2 + 2b2) x + (a2 - b2) = 0 (a - b) (a + b) x2 - (2a2 + 2b2) x + (a - b) (a + b) = 0 [(a - b) x - (a + b)] [(a + b) x - (a - b)] = 0 x = (a+b)/(a-b) or x = (a-b)/(a+b) If a=b≠0, The original equation becomes 4a2 (x2 - x + 1) = 4a2 (x2 + x + 1) 8a2 x = 0 x = 0 If a=b=0, The equation becomes 0 = 0 and so every real x satisfies the equation. If a=-b≠0, The original equation becomes 4a2 (x2 - x + 1) = 4a2 (x2 + x + 1) 8a2 x = 0 x = 0 (2) In order for easy typing, let u = x1/3, v = y1/3, w = z1/3, Then the question becomes "If u+v+w=0, show that (u3 + v3 + w3)3 = 27 u3 v3 w3" Since u+v+w=0, by taking cube on both sides, (u+v+w)3 = 0 u3 + v3 + w3 + 3u2v + 3uv2 + 3v2w + 3vw2 + 3u2w + 3uw2 + 6uvw = 0【Check by yourself】 (u3 + v3 + w3) + (3u2v + 3uv2) + (3v2w + 3vw2) + (3u2w + 3uw2) + 6uvw = 0 (u3 + v3 + w3) + (3u2v + 3uv2 + 3uvw) + (3v2w + 3vw2 + 3uvw) + (3u2w + 3uw2 + 3uvw) + 6uvw - 9uvw = 0【add 3uvw in each bracket then -9uvw at the end】 (u3 + v3 + w3) + 3uv (u + v + w) + 3vw (v + w + u) + 3uw (u + w + v) -3uvw = 0 (u3 + v3 + w3) + 3uv (0) + 3vw (0) + 3uw (0) -3uvw = 0【Since u+v+w=0】 u3 + v3 + w3 = 3uvw Finally, taking cube again on both sides, we have the answer: (u3 + v3 + w3)3 = 27u3 v3 w3 Hope it helps! ^^ 2006-12-21 14:05:31 補充： (2)So changing back into x, y, z,(x^(1/3)3＋y^(1/3)3＋z^(1/3)3)3＝27x^(1/3)3 y^(1/3)3 z^(1/3)3x＋y＋z＝27 x y z 2006-12-21 14:08:44 補充： Sorry, some wrong typing in the supplement:(2)So changing back into x, y, z,(x^(1/3)3＋y^(1/3)3＋z^(1/3)3)3＝27x^(1/3)3 y^(1/3)3 z^(1/3)3(x＋y＋z)3＝27xyz
其他解答:
Challenge!
發問:
Please help! 1. Solve x if (3a^2 + b^2) (x^2 - x + 1) = (3b^2 + a^2) (x^2 + x + 1). 2. If x^1/3 + y^1/3 + z^1/3 = 0, show that (x+y+z)^3 = 27xyz Thanks!
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最佳解答:(1) If a≠b and a≠-b, (3a2 + b2) (x2 - x + 1) = (3b2 + a2) (x2 + x + 1) (3a2 + b2) x2 - (3a2 + b2) x + (3a2 + b2) = (3b2 + a2) x2 + (3b2 + a2) x + (3b2 + a2) (2a2 - 2b2) x2 - (4a2 + 4b2) x + (2a2 - 2b2) = 0 (a2 - b2) x2 - (2a2 + 2b2) x + (a2 - b2) = 0 (a - b) (a + b) x2 - (2a2 + 2b2) x + (a - b) (a + b) = 0 [(a - b) x - (a + b)] [(a + b) x - (a - b)] = 0 x = (a+b)/(a-b) or x = (a-b)/(a+b) If a=b≠0, The original equation becomes 4a2 (x2 - x + 1) = 4a2 (x2 + x + 1) 8a2 x = 0 x = 0 If a=b=0, The equation becomes 0 = 0 and so every real x satisfies the equation. If a=-b≠0, The original equation becomes 4a2 (x2 - x + 1) = 4a2 (x2 + x + 1) 8a2 x = 0 x = 0 (2) In order for easy typing, let u = x1/3, v = y1/3, w = z1/3, Then the question becomes "If u+v+w=0, show that (u3 + v3 + w3)3 = 27 u3 v3 w3" Since u+v+w=0, by taking cube on both sides, (u+v+w)3 = 0 u3 + v3 + w3 + 3u2v + 3uv2 + 3v2w + 3vw2 + 3u2w + 3uw2 + 6uvw = 0【Check by yourself】 (u3 + v3 + w3) + (3u2v + 3uv2) + (3v2w + 3vw2) + (3u2w + 3uw2) + 6uvw = 0 (u3 + v3 + w3) + (3u2v + 3uv2 + 3uvw) + (3v2w + 3vw2 + 3uvw) + (3u2w + 3uw2 + 3uvw) + 6uvw - 9uvw = 0【add 3uvw in each bracket then -9uvw at the end】 (u3 + v3 + w3) + 3uv (u + v + w) + 3vw (v + w + u) + 3uw (u + w + v) -3uvw = 0 (u3 + v3 + w3) + 3uv (0) + 3vw (0) + 3uw (0) -3uvw = 0【Since u+v+w=0】 u3 + v3 + w3 = 3uvw Finally, taking cube again on both sides, we have the answer: (u3 + v3 + w3)3 = 27u3 v3 w3 Hope it helps! ^^ 2006-12-21 14:05:31 補充： (2)So changing back into x, y, z,(x^(1/3)3＋y^(1/3)3＋z^(1/3)3)3＝27x^(1/3)3 y^(1/3)3 z^(1/3)3x＋y＋z＝27 x y z 2006-12-21 14:08:44 補充： Sorry, some wrong typing in the supplement:(2)So changing back into x, y, z,(x^(1/3)3＋y^(1/3)3＋z^(1/3)3)3＝27x^(1/3)3 y^(1/3)3 z^(1/3)3(x＋y＋z)3＝27xyz
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