Phân tích đa thức thành nhân tử P = $a^{3}$ (b – c) + $b^{3}$ (c – a) + $c^{3}$ (a – b)

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1. Đáp án :

   `P=(b-c)(a-b)(a-c)(a+b+c)`

   Giải thích các bước giải :

   `P=a^3(b-c)+b^3(c-a)+c^3(a-b)`
   `<=>P=a^3(b-c)+b^3(c-b+b-a)+c^3(a-b)`
   `<=>P=a^3(b-c)+b^3[-(b-c)-(a-b)]+c^3(a-b)`
   `<=>P=a^3(b-c)-b^3[(b-c)+(a-b)]+c^3(a-b)`
   `<=>P=a^3(b-c)-b^3(b-c)-b^3(a-b)+c^3(a-b)`
   `<=>P=(b-c)(a^3-b^3)+(a-b)(c^3-b^3)`
   `<=>P=(b-c)(a^3-b^3)-(a-b)(b^3-c^3)`
   `<=>P=(b-c)(a-b)(a^2+ab+b^2)-(a-b)(b-c)(b^2+bc+c^2)`
   `<=>P=(b-c)(a-b)[(a^2+ab+b^2)-(b^2+bc+c^2)]`
   `<=>P=(b-c)(a-b)[a^2+ab+b^2-b^2-bc-c^2]`
   `<=>P=(b-c)(a-b)[a^2+ab-bc-c^2]`
   `<=>P=(b-c)(a-b)[(a^2-c^2)+(ab-bc)]`
   `<=>P=(b-c)(a-b)[(a-c)(a+c)+b(a-c)]`
   `<=>P=(b-c)(a-b)(a-c)(a+b+c)`
   Vậy : `P=(b-c)(a-b)(a-c)(a+b+c)`

   Trả lời

2.  a³(b – c) + b³(c – a) + c³(a – b)

   = a³(b – c) + b³(c – a + b – b) + c³(a – b)

   = a³(b – c) + b³[ (c – b) + (b – a) ] + c³(a – b)

   = a³(b – c) + b³(c – b) + b³(b – a) + c³(a – b)

   = a³(b – c) – b³(b – c) – b³(a – b) + c³(a – b)

   = [ a³(b – c) – b³(b – c) ] + [ c³(a – b) – b³(a – b) ] 

   = (b – c)(a³ – b³) + (a – b)(c³ – b³)

   = (b – c)(a – b)(a² + ab + b²) + (a – b)(c – b)(c² + cb + b²)

   = (b – c)(a – b)(a² + ab + b²) – (a – b)(b – c)(c² + cb + b²)

   = (b – c)(a – b)[ (a² + ab + b²) – (c² + cb + b²) ]

   = (b – c)(a – b)(a² + ab + b² – c² – cb – b²)

   = (b – c)(a – b)(a² + ab – c² – cb)

   = (b – c)(a – b)[ (a² – c²) + (ab + cb) ]

   = (b – c)(a – b)[ (a – c)(a + c) + b(a + c) ]

   = (b – c)(a – b)[ (a – c)(a + c + b) ]

   = (b – c)(a – b)(a – c)(a + c + b)

   Vậy a³(b – c) + b³(c – a) + c³(a – b) = (b – c)(a – b)(a – c)(a + c + b)

   Trả lời