Question

Factorice:(a-b) ^3 +(b-c) ^3 +(c-a) ^3​

Answer 1

Answer:

$here \: we \: have \: to \: use \: the \: formula \: \\ (x - y) ^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) \\ \\ \\ = (a - b) ^{3} + (b - c) ^{3} + (c - a) ^{3} \\ \\ = {a}^{3} - {b}^{3} - 3ab(a - b) \\ \\ = {a}^{3} - {b}^{3} - 3 {a}^{2} b + 3a {b}^{2} ......i \\ \\ = {b}^{3} - {c}^{3} - 3bc(b - c) \\ \\ = {b}^{3} - {c}^{3} - 3 {b}^{2} c + 3b {c}^{2} .......ii \\ \\ = {c}^{3} - {a}^{3} - 3ca(c - a) \\ \\ = {c}^{3} - {a}^{3} - 3 {c}^{2} a + 3c {a}^{2} .......iii \\ \\ \\ now \: add \: i + ii + iii \\ \\ = ( {a}^{3} - {b}^{3} - 3 {a}^{2} b + 3a {b}^{2} ) + \\ ( {b}^{3} - {c}^{3} - 3 {b}^{2} c + 3b {c}^{2} ) + \\ ( {c}^{3} - {a}^{3} - 3 {c}^{2} a + 3c {a}^{2} ) \\ \\ = {a}^{3} - {a}^{3} - {b}^{3} + {b}^{3} - {c}^{3} + {c}^{3} - 3 {a}^{2} b + 3a {b}^{2} - 3 {b}^{2} c + 3b {c}^{2} - 3 {c}^{2} a + 3c {a}^{2} \\ \\ = - 3 {a}^{2} b + 3a {b}^{2} - 3 {b}^{2} c + 3b {c}^{2} - 3 {c}^{2} a + 3c {a}^{2}$

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