Proof. Let a, b, c ∈ R such that abc ∈
√
I ⇒ (abc)
n ∈
√
I for some n ∈ N.
That is, abc ∈ I or (abc)
2 ∈ I or (abc)
3 ∈ I.
Case 1. If abc ∈ I, then we are done, since I is 2-absorbing and I ⊂
√
I
Case 2. If (abc)
2 ∈ I, then a
2
b
2
c
2 ∈ I ⇒ a
2
b
2 ∈ I or b
2
c
2 ∈ I or a
2
c
2 ∈ I. So
that (ab)
2 ∈ I or (bc)
2 ∈ I or (ac)
2 ∈ I ⇒ ab ∈
√
I or bc ∈
√
I or ac ∈
√
I.
Case 3. If (abc)
3 ∈ I, then (abc)
2 ∈ I and the result follows from case 2.
Proof. Let a, b, c ∈ R such that abc ∈√I ⇒ (abc)n ∈√I for some n ∈ N.That is, abc ∈ I or (abc)2 ∈ I or (abc)3 ∈ I.Case 1. If abc ∈ I, then we are done, since I is 2-absorbing and I ⊂√ICase 2. If (abc)2 ∈ I, then a2b2c2 ∈ I ⇒ a2b2 ∈ I or b2c2 ∈ I or a2c2 ∈ I. Sothat (ab)2 ∈ I or (bc)2 ∈ I or (ac)2 ∈ I ⇒ ab ∈√I or bc ∈√I or ac ∈√I.Case 3. If (abc)3 ∈ I, then (abc)2 ∈ I and the result follows from case 2.
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