Zimmerman Logan
05/24/2024 · Middle School
Factorizar por el método anterior, si es viamente: 1. \( a^{3}+3 a^{2}+3 a+1 \) 2. \( 27-27 x+9 x^{2}-x^{3} \) 3. \( m^{3}+3 m^{2} n+3 m n^{2}+n^{3} \) 4. \( 1+3 a^{2}-3 a-a^{3} \) 5. \( 8+12 a^{2}+6 a^{4}+a^{6} \) 6. \( 125 x^{3}+1+75 x^{2}+15 x \) 7. \( 8 a^{3}-36 a^{2} b+54 a b^{2}-27 b^{3} \) 8. \( 27 m^{3}+108 m^{2} n+144 m n^{2}+64 n^{3} \) 9. \( x^{3}-3 x^{2}+3 x+1 \) 1. \( 1+12 a^{2} b-6 a b-8 a^{3} b^{3} \) 11. \( 125 a^{3}+150 a^{2} b+60 a b^{2}+8 b^{3} \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
Vamos a factorizar cada uno de los polinomios que has proporcionado.
1. **\( a^{3}+3 a^{2}+3 a+1 \)**
Este polinomio se puede reconocer como un cubo perfecto:
\[
a^{3}+3 a^{2}+3 a+1 = (a+1)^{3}
\]
2. **\( 27-27 x+9 x^{2}-x^{3} \)**
Reordenando los términos, tenemos:
\[
-x^{3}+9 x^{2}-27 x+27 = -(x^{3}-9 x^{2}+27 x-27)
\]
Factorizando el polinomio:
\[
-(x-3)^{3}
\]
3. **\( m^{3}+3 m^{2} n+3 m n^{2}+n^{3} \)**
Este es otro cubo perfecto:
\[
m^{3}+3 m^{2} n+3 m n^{2}+n^{3} = (m+n)^{3}
\]
4. **\( 1+3 a^{2}-3 a-a^{3} \)**
Reordenando los términos:
\[
-a^{3}+3 a^{2}-3 a+1 = -(a^{3}-3 a^{2}+3 a-1)
\]
Este es un cubo perfecto:
\[
-(a-1)^{3}
\]
5. **\( 8+12 a^{2}+6 a^{4}+a^{6} \)**
Reordenando:
\[
a^{6}+6 a^{4}+12 a^{2}+8 = (a^{2}+2)^{3}
\]
6. **\( 125 x^{3}+1+75 x^{2}+15 x \)**
Reordenando:
\[
125 x^{3}+75 x^{2}+15 x+1 = (5x+1)^{3}
\]
7. **\( 8 a^{3}-36 a^{2} b+54 a b^{2}-27 b^{3} \)**
Reordenando:
\[
8 a^{3}-36 a^{2} b+54 a b^{2}-27 b^{3} = (2a-3b)^{3}
\]
8. **\( 27 m^{3}+108 m^{2} n+144 m n^{2}+64 n^{3} \)**
Este es un cubo perfecto:
\[
(3m+4n)^{3}
\]
9. **\( x^{3}-3 x^{2}+3 x+1 \)**
Este polinomio se puede reconocer como un cubo perfecto:
\[
(x-1)^{3}
\]
10. **\( 1+12 a^{2} b-6 a b-8 a^{3} b^{3} \)**
Reordenando:
\[
-8 a^{3} b^{3}+12 a^{2} b-6 a+1 = (1-2ab)^{3}
\]
11. **\( 125 a^{3}+150 a^{2} b+60 a b^{2}+8 b^{3} \)**
Este es un cubo perfecto:
\[
(5a+2b)^{3}
\]
Espero que estas factorizaciones sean útiles. Si necesitas más ayuda, no dudes en preguntar.
Quick Answer
1. \( a^{3}+3 a^{2}+3 a+1 = (a+1)^{3} \)
2. \( 27-27 x+9 x^{2}-x^{3} = -(x-3)^{3} \)
3. \( m^{3}+3 m^{2} n+3 m n^{2}+n^{3} = (m+n)^{3} \)
4. \( 1+3 a^{2}-3 a-a^{3} = -(a-1)^{3} \)
5. \( 8+12 a^{2}+6 a^{4}+a^{6} = (a^{2}+2)^{3} \)
6. \( 125 x^{3}+1+75 x^{2}+15 x = (5x+1)^{3} \)
7. \( 8 a^{3}-36 a^{2} b+54 a b^{2}-27 b^{3} = (2a-3b)^{3} \)
8. \( 27 m^{3}+108 m^{2} n+144 m n^{2}+64 n^{3} = (3m+4n)^{3} \)
9. \( x^{3}-3 x^{2}+3 x+1 = (x-1)^{3} \)
10. \( 1+12 a^{2} b-6 a b-8 a^{3} b^{3} = (1-2ab)^{3} \)
11. \( 125 a^{3}+150 a^{2} b+60 a b^{2}+8 b^{3} = (5a+2b)^{3} \)
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