(b-c)^3+(c-a)^3+(a-b)^3 Question (b-c)^3+(c-a)^3+(a-b)^3 in progress 0 Môn Toán Nick 4 years 2020-11-01T12:05:10+00:00 2020-11-01T12:05:10+00:00 2 Answers 49 views 0
Answers ( )
$\begin{array}{l}(a -b)^3 + (b-c)^3 + (c-a)^3\\ = (a – b + b – c)[(a-b)^2 – (a-b)(b-c) + (b-c)^2] + (c-a)^3\\ = (a – c)(a^2 – 2ab + b^2 – ab + ac + b^2 – bc + b^2 – 2bc + c^2) – (a – c)^3\\ = (a-c)(a^2 – 3ab + 3b^2 + ac – 3bc + c^2 – (a^2 – 2ac + c^2))\\ = (a-c)(-3ab + 3b^2 + 3ac – 3bc)\\ = 3(a-c)[b(b-a) -c(b – a)]\\ = 3(a-c)(b-a)(b-c)\end{array}$
Giải thích các bước giải:
`(b – c)^3 + (c – a)^3 + (a – b)^3`
`=(b^3 – 3b^2c + 3bc^2 – c^3) + (c^3 – 3c^2a + 3ca^2 – a^3) + (a^3 – 3a^2b + 3ab^2 – b^3)`
`=b^3 – 3b^2c + 3bc^2 – c^3 + c^3 – 3c^2a + 3ca^2 – a^3 + a^3 – 3a^2b + 3ab^2 – b^3`
`=-3b^2c + 3bc^2 – 3c^2a + 3ca^2 – 3a^2b + 3ab^2`
`=-3(b^2c – bc^2 + c^2a – ca^2 + a^2b – ab^2)`
`=-3[(a^2b – ab^2) + (c^2a – bc^2) – (a^2c – b^2c)]`
`=-3[ab(a – b) + c^2(a – b) – c(a^2 – b^2)]`
`=-3[ab(a – b) + c^2(a – b) – c(a + b)(a – b)]`
`=-3(a-b)[ab + c^2 – c(a + b)]`
`=-3(a – b)(ab + c^2 – ca – cb)`
`=-3(a – b)[(ab – ca) – (cb – c^2)]`
`=-3(a – b)[a(b – c) – c(b – c)]`
`=-3(a – b)(b – c)(a – c)`