Ans:

$-$ Ta có:

$a^3 - 7a - 6$

$= a^3 - a - 6a - 6$

$= (a^3 - a) - (6a + 6)$

$= a(a^2 - 1) - 6(a + 1)$

$= a(a - 1)(a + 1) - 6(a + 1)$

$= (a + 1)[a(a - 1) - 6]$

$= (a + 1)(a^2 - a - 6)$

$= (a + 1)(a^2 - 3a + 2a - 6)$

$= (a + 1)[a(a - 3) + 2(a - 3)]$

$= (a + 1)(a - 3)(a + 2)$

⇒ $(a + 1)(a + 2)(a - 3)$

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