prove that: (a2 - b2)3 + (b2-c2)3+ (c2-a2)3 = 3 (a+b) (b+c) (c+a) (a-b) (b-c) (c-a).

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27-07-2014
Mathematics

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prove that: (a2 - b2)3 + (b2-c2)3+ (c2-a2)3 = 3 (a+b) (b+c) (c+a) (a-b) (b-c) (c-a).

Answer :

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kayt91 kayt91 · Answer 1 · 2014-07-27T19:30:38-04:00

[tex] (a^2 - b^2)^3 + (b^2 - c^2)3 + (c^2 - a^2)^3 = 3(a + b)(b +c)(c + a)(a - b)(b - c)(c - a)[/tex]
[tex] (a^2 - b^2)^3 + (b^2 - c^2)3 + (c^2 - a^2)^3 = 3(a + b)(a - b)(b + c)(b - c)(c + a)(c - a) [/tex]
[tex] (a^2 - b^2)^3 + (b^2 - c^2)3 + (c^2 - a^2)^3 = 3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2) [/tex]

[tex] (a^2 - b^2)^3 = (a^2 - b2)(a^2 - b^2)(a^2-b^2) = a^6 - 3a^4b^2 + 3a^2b^4 - b^6 [/tex]
[tex] (b^2 - c^2)^3 = (b^2 - c^2)(b^2 - c^2)(b^2 - c^2) = b^6 - 3^4c^2 + 3b^2c^4 - c^6) [/tex]
[tex] (c^2 - a^2)^3 = (c^2 - a^2)(c^2 - a^2)(c^2 - a^2) = c^6 - 3a^2c^4 + 3a^4c^2 - a^6 [/tex]

[tex] a^6 - 3a^4b^2 + 3a^2b^4 - b^6 + b^6 - 3b^4c^2 + 3b^2c^4 - c^6 + c^6 - 3a^2c^4 + 3a^4c^2 - a^6 [/tex]
[tex] -3a^4b^2 + 3a^2b^4 - 3b^4c^2 + 3b^2c^4 - 3a^2c^4 + 3a^4c^2 [/tex]
[tex] 3(-a^4b^2 + a^2b^4 - b^4c^2 + b2^c^4 - a^2c^4 + a^4c^2) [/tex]

[tex] 3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2) = 3(-a^4b^2 + a^2b^4 - b^4c^2 + b^2c^4 - a^2c^4 + a^4c^2) [/tex]

[tex] 3(-a^4b^2 + a^2b^4 - b^4c^2 + b^2c^4 - a^2c^4 + a^4c^2) = 3(-a^4b^2 + a^2b^4 - b^4c^2 + b^2c^4 - a^2c^4 + a^4c^2 [/tex]

Аноним Аноним · Answer 2 · 2014-07-27T19:52:16-04:00

[tex]L=(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=(*)\\\\(a^2-b^2)^3=a^6-3a^4b^2+3a^2b^4-b^6\\\\(b^2-c^2)^3=b^6-3b^4c^2+3b^3c^4-c^6\\\\(c^2-a^2)^3=c^6-3c^4a^2+3c^2a^4-a^6\\\\(*)=a^6-3a^4b^2+3a^2b^4-b^6+b^6-3b^4c^2+3b^2c^4-c^6+c^6-\dots\\\dots-3c^4a^2+3c^2a^4-a^6\\\\=-3a^4b^2+3a^2b^4-3b^4c^2+3b^2c^4-3c^4a^2+3c^2a^4\\\\=3(-a^4b^2+a^2b^4-b^4c^2+b^2c^4-a^2c^4+a^4c^2)[/tex]

[tex]R=3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a)\\\\=3(a^2-b^2)(b^2-c^2)(c^2-a^2)\\\\=3(a^2b^2-a^2c^2-b^4+b^2c^2)(c^2-a^2)\\\\=3(a^2b^2c^2-a^4b^2-a^2c^4+a^4c^2-b^4c^2+a^2b^4+b^2c^4-a^2b^2c^2)\\\\=3(-a^4b^2+a^2b^4-b^4c^2+b^2c^4-a^2c^4+a^4c^2)\\\\L=R[/tex]