Wagner Marsh
10/10/2023 · Elementary School
EJERCICIO 103 Descomponer en 2 factore \( \begin{array}{ll}\text { 1. } 1+a^{3}, & \text { 7. } y^{3}-1 \\ \text { 2. } 1-a^{3} & \text { 8. } 8 x^{3}-1 \\ \text { 3. } x^{3}+y^{3}, & \text { 9. } 1-8 x^{3} \\ \text { 4. } m^{3}-n^{3} & \text { 10. } x^{3}-27 \\ \text { 5. } a^{3}-1 . & \text { 11. } a^{3}+27 \\ \text { 6. } y^{3}+1 & \text { 12. } 8 x^{3}+y^{3}\end{array} \)
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Para descomponer las expresiones en dos factores, utilizaremos las identidades de factorización para sumas y restas de cubos, así como otras factorizaciones relevantes. Aquí están las descomposiciones:
1. \( 1 + a^3 = (1 + a)(1 - a + a^2) \)
2. \( 1 - a^3 = (1 - a)(1 + a + a^2) \)
3. \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \)
4. \( m^3 - n^3 = (m - n)(m^2 + mn + n^2) \)
5. \( a^3 - 1 = (a - 1)(a^2 + a + 1) \)
6. \( y^3 + 1 = (y + 1)(y^2 - y + 1) \)
7. \( y^3 - 1 = (y - 1)(y^2 + y + 1) \)
8. \( 8x^3 - 1 = (2x - 1)(4x^2 + 2x + 1) \)
9. \( 1 - 8x^3 = (1 - 2x)(1 + 2x + 4x^2) \)
10. \( x^3 - 27 = (x - 3)(x^2 + 3x + 9) \)
11. \( a^3 + 27 = (a + 3)(a^2 - 3a + 9) \)
12. \( 8x^3 + y^3 = (2x + y)(4x^2 - 2xy + y^2) \)
Estas son las factorizaciones de las expresiones dadas.
Quick Answer
1. \( 1 + a^3 = (1 + a)(1 - a + a^2) \)
2. \( 1 - a^3 = (1 - a)(1 + a + a^2) \)
3. \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \)
4. \( m^3 - n^3 = (m - n)(m^2 + mn + n^2) \)
5. \( a^3 - 1 = (a - 1)(a^2 + a + 1) \)
6. \( y^3 + 1 = (y + 1)(y^2 - y + 1) \)
7. \( y^3 - 1 = (y - 1)(y^2 + y + 1) \)
8. \( 8x^3 - 1 = (2x - 1)(4x^2 + 2x + 1) \)
9. \( 1 - 8x^3 = (1 - 2x)(1 + 2x + 4x^2) \)
10. \( x^3 - 27 = (x - 3)(x^2 + 3x + 9) \)
11. \( a^3 + 27 = (a + 3)(a^2 - 3a + 9) \)
12. \( 8x^3 + y^3 = (2x + y)(4x^2 - 2xy + y^2) \)
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