Question

answer the question pls..​

Answer 1

Answer:

$we \: \: should \: \: write \: \: this \: \: as = > \\ \\ (x - a) + (x - b) + (x - c) = 0 \\ \\ = > (x - a) + (x- b) = - (x - c) \\ \\ = > cube \: \: both \: \: side - \\ \\ = > {((x - a)}^{} + {(x- b))}^{3} = - {(x - c)}^{3} \\ \\ = > {(x - a)}^{3} + {(x - b)}^{3} + 3(x - a)(x - b)(x - a + x - b) = - {(x - c)}^{3} \\ \\ = > {(x - a)}^{3} + {(x - b)}^{3} + {(x - c)}^{3} = - 3(x - a)(x - b)(2x - a - b) \\ \\ = - 3(x - a)(x - b)(2x - (a + b)) \\ \\ = - 3(x - a)(x - b)(2x - (3x - c)) \: \: \: \: \: \: by \: using \: \: given \: \: equation\\ \\ = - 3(x - a)(x - b)( - x + c) \\ \\ = 3(x - a)(x - b)(x - c)$

So , option (b) is correct...

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