Here,
First expression = 3a3 - 27a
= 3a(a2 - 9)
= 3a (a2 - 32)
= 3a (a+3) (a-3)
Second expression = a3 + 27
= a3 + 33
= (a + 3) (a2 - 3a + 9)
Therefore, HCF = (a + 3).
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Find the HCF of 3a3 - 27a and a3 + 27.
Here,
First expression = 3a3 - 27a
= 3a(a2 - 9)
= 3a (a2 - 32)
= 3a (a+3) (a-3)
Second expression = a3 + 27
= a3 + 33
= (a + 3) (a2 - 3a + 9)
Therefore, HCF = (a + 3).
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