Factorise $4 a b c {(a + b - c)}^{2} - 2 a b (a + b - c)$ .

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Solution

Step I - Here $2 a b (a + b - c)$ is the common factor from each term.
So we get, $4 a b c {(a + b - c)}^{2} - 2 a b (a + b - c)$ = $2 a b (a + b - c)$ $[2 c (a + b - c) - 1]$
= $2 a b (a + b - c)$ $(2 a c + 2 b c - 2 c^{2} - 1)$
Hence, the factors of $4 a b c {(a + b - c)}^{2} - 2 a b (a + b - c)$ = $2 a b (a + b - c)$ $(2 a b + 2 b c - 2 c^{2} - 1)$ .

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Question 92 (xxvi)

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