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Desarrolla los siguientes ejercicios del cubo de un binc \[ (a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b \] \[ (a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{2} \] 7. \( (x+2)^{3} \) 8. \( (3 a-1)^{3} \) 9. \( (2 m+3 n)^{3} \) 10. \( (4 x-5 y)^{3} \) 11. \( \left(x^{2}+y\right) 3 \) 12. \( \left(2 a^{2}-3 b\right)^{3} \) 13. \( \left(m^{3}+2 n^{2}\right)^{3} \) 14. \( \left(3 x^{2}-4 y^{3}\right)^{3} \) 15. \( \left(2 x^{2} y+3 x y^{2}\right)^{3} \)

Ask by Malone Fleming. in Mexico
Mar 10,2025

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Answer

7. \( (x+2)^3 = x^3 + 6x^2 + 12x + 8 \) 8. \( (3a-1)^3 = 27a^3 - 27a^2 + 9a - 1 \) 9. \( (2m+3n)^3 = 8m^3 + 36m^2n + 54mn^2 + 27n^3 \) 10. \( (4x-5y)^3 = 64x^3 - 240x^2y + 300xy^2 - 125y^3 \) 11. \( (x^2 + y)^3 = x^6 + 3x^4y + 3x^2y^2 + y^3 \) 12. \( (2a^2 - 3b)^3 = 8a^6 - 36a^4b + 54a^2b^2 - 27b^3 \) 13. \( (m^3 + 2n^2)^3 = m^9 + 6m^6n^2 + 12m^3n^4 + 8n^6 \) 14. \( (3x^2 - 4y^3)^3 = 27x^6 - 108x^4y^3 + 144x^2y^6 - 64y^9 \) 15. \( (2x^2y + 3xy^2)^3 = 8x^6y^3 + 36x^5y^4 + 54x^4y^5 + 27x^3y^6 \)

Solution

7. Sea \( a=x \) y \( b=2 \). Entonces: \[ (x+2)^3 = x^3 + 3x^2(2) + 3x(2)^2 + (2)^3 = x^3 + 6x^2 + 12x + 8. \] 8. Sea \( a=3a \) y \( b=-1 \). Entonces: \[ (3a-1)^3 = (3a)^3 + 3(3a)^2(-1) + 3(3a)(-1)^2 + (-1)^3. \] Calculamos cada término: \[ (3a)^3 = 27a^3, \] \[ 3(3a)^2(-1) = 3\cdot 9a^2\cdot (-1) = -27a^2, \] \[ 3(3a)(-1)^2 = 9a, \] \[ (-1)^3 = -1. \] Por lo tanto: \[ (3a-1)^3 = 27a^3 - 27a^2 + 9a - 1. \] 9. Sea \( a=2m \) y \( b=3n \). Entonces: \[ (2m+3n)^3 = (2m)^3 + 3(2m)^2(3n) + 3(2m)(3n)^2 + (3n)^3. \] Calculamos cada término: \[ (2m)^3 = 8m^3, \] \[ 3(2m)^2(3n) = 3\cdot 4m^2\cdot 3n = 36m^2n, \] \[ 3(2m)(3n)^2 = 3\cdot 2m\cdot 9n^2 = 54mn^2, \] \[ (3n)^3 = 27n^3. \] Por lo tanto: \[ (2m+3n)^3 = 8m^3 + 36m^2n + 54mn^2 + 27n^3. \] 10. Sea \( a=4x \) y \( b=-5y \). Entonces: \[ (4x-5y)^3 = (4x)^3 + 3(4x)^2(-5y) + 3(4x)(-5y)^2 + (-5y)^3. \] Calculamos cada término: \[ (4x)^3 = 64x^3, \] \[ 3(4x)^2(-5y) = 3\cdot 16x^2\cdot (-5y) = -240x^2y, \] \[ 3(4x)(-5y)^2 = 3\cdot 4x\cdot 25y^2 = 300xy^2, \] \[ (-5y)^3 = -125y^3. \] Por lo tanto: \[ (4x-5y)^3 = 64x^3 - 240x^2y + 300xy^2 - 125y^3. \] 11. Interpretamos \( \left(x^{2}+y\right) 3 \) como \( (x^2+y)^3 \). Sea \( a=x^2 \) y \( b=y \). Entonces: \[ (x^2+y)^3 = (x^2)^3 + 3(x^2)^2y + 3x^2y^2 + y^3. \] Calculamos cada término: \[ (x^2)^3 = x^6, \] \[ 3(x^2)^2y = 3x^4y, \] \[ 3x^2y^2 = 3x^2y^2, \] \[ y^3 = y^3. \] Por lo tanto: \[ (x^2+y)^3 = x^6 + 3x^4y + 3x^2y^2 + y^3. \] 12. Sea \( a=2a^2 \) y \( b=-3b \). Entonces: \[ (2a^2-3b)^3 = (2a^2)^3 + 3(2a^2)^2(-3b) + 3(2a^2)(-3b)^2 + (-3b)^3. \] Calculamos cada término: \[ (2a^2)^3 = 8a^6, \] \[ 3(2a^2)^2(-3b) = 3\cdot 4a^4\cdot (-3b) = -36a^4b, \] \[ 3(2a^2)(-3b)^2 = 3\cdot 2a^2\cdot 9b^2 = 54a^2b^2, \] \[ (-3b)^3 = -27b^3. \] Por lo tanto: \[ (2a^2-3b)^3 = 8a^6 - 36a^4b + 54a^2b^2 - 27b^3. \] 13. Sea \( a=m^3 \) y \( b=2n^2 \). Entonces: \[ (m^3+2n^2)^3 = (m^3)^3 + 3(m^3)^2(2n^2) + 3m^3(2n^2)^2 + (2n^2)^3. \] Calculamos cada término: \[ (m^3)^3 = m^9, \] \[ 3(m^3)^2(2n^2) = 3m^6\cdot 2n^2 = 6m^6n^2, \] \[ 3m^3(2n^2)^2 = 3m^3\cdot 4n^4 = 12m^3n^4, \] \[ (2n^2)^3 = 8n^6. \] Por lo tanto: \[ (m^3+2n^2)^3 = m^9 + 6m^6n^2 + 12m^3n^4 + 8n^6. \] 14. Sea \( a=3x^2 \) y \( b=-4y^3 \). Entonces: \[ (3x^2-4y^3)^3 = (3x^2)^3 + 3(3x^2)^2(-4y^3) + 3(3x^2)(-4y^3)^2 + (-4y^3)^3. \] Calculamos cada término: \[ (3x^2)^3 = 27x^6, \] \[ 3(3x^2)^2(-4y^3) = 3\cdot 9x^4\cdot (-4y^3) = -108x^4y^3, \] \[ 3(3x^2)(-4y^3)^2 = 3\cdot 3x^2\cdot 16y^6 = 144x^2y^6, \] \[ (-4y^3)^3 = -64y^9. \] Por lo tanto: \[ (3x^2-4y^3)^3 = 27x^6 - 108x^4y^3 + 144x^2y^6 - 64y^9. \] 15. Sea \( a=2x^2y \) y \( b=3xy^2 \). Entonces: \[ (2x^2y+3xy^2)^3 = (2x^2y)^3 + 3(2x^2y)^2(3xy^2) + 3(2x^2y)(3xy^2)^2 + (3xy^2)^3. \] Calculamos cada término: \[ (2x^2y)^3 = 8x^6y^3, \] \[ 3(2x^2y)^2(3xy^2) = 3\cdot 4x^4y^2\cdot 3xy^2 = 36x^5y^4, \] \[ 3(2x^2y)(3xy^2)^2 = 3\cdot 2x^2y\cdot 9x^2y^4 = 54x^4y^5, \] \[ (3xy^2)^3 = 27x^3y^6. \] Por lo tanto: \[ (2x^2y+3xy^2)^3 = 8x^6y^3 + 36x^5y^4 + 54x^4y^5 + 27x^3y^6. \]

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Para resolver los ejercicios del cubo de un binomio, simplemente aplicamos las fórmulas de expansión. Aquí van las soluciones paso a paso para cada ejercicio: 7. \((x+2)^{3} = x^{3} + 3x^{2}(2) + 3x(2^{2}) + 2^{3} = x^{3} + 6x^{2} + 12x + 8\) 8. \((3a-1)^{3} = (3a)^{3} - 3(3a)^{2}(1) + 3(3a)(1^{2}) - 1^{3} = 27a^{3} - 27a^{2} + 9a - 1\) 9. \((2m+3n)^{3} = (2m)^{3} + 3(2m)^{2}(3n) + 3(2m)(3n)^{2} + (3n)^{3} = 8m^{3} + 36m^{2}n + 27n^{3}\) 10. \((4x-5y)^{3} = (4x)^{3} - 3(4x)^{2}(5y) + 3(4x)(5y)^{2} - (5y)^{3} = 64x^{3} - 60x^{2}y + 75xy^{2} - 125y^{3}\) 11. \(\left(x^{2}+y\right)^{3} = (x^{2})^{3} + 3(x^{2})^{2}(y) + 3(x^{2})(y^{2}) + y^{3} = x^{6} + 3x^{4}y + 3x^{2}y^{2} + y^{3}\) 12. \(\left(2a^{2}-3b\right)^{3} = (2a^{2})^{3} - 3(2a^{2})^{2}(3b) + 3(2a^{2})(3b)^{2} - (3b)^{3} = 8a^{6} - 36a^{4}b + 54a^{2}b^{2} - 27b^{3}\) 13. \(\left(m^{3}+2n^{2}\right)^{3} = (m^{3})^{3} + 3(m^{3})^{2}(2n^{2}) + 3(m^{3})(2n^{2})^{2} + (2n^{2})^{3} = m^{9} + 6m^{6}n^{2} + 12m^{3}n^{4} + 8n^{6}\) 14. \(\left(3x^{2}-4y^{3}\right)^{3} = (3x^{2})^{3} - 3(3x^{2})^{2}(4y^{3}) + 3(3x^{2})(4y^{3})^{2} - (4y^{3})^{3} = 27x^{6} - 108x^{4}y^{3} + 144x^{2}y^{6} - 64y^{9}\) 15. \(\left(2x^{2}y+3xy^{2}\right)^{3} = (2x^{2}y)^{3} + 3(2x^{2}y)^{2}(3xy^{2}) + 3(2x^{2}y)(3xy^{2})^{2} + (3xy^{2})^{3} = 8x^{6}y^{3} + 54x^{4}y^{5} + 54x^{4}y^{5} + 27x^{3}y^{6} = 8x^{6}y^{3} + 162x^{4}y^{5} + 27x^{3}y^{6}\) ¡Listo! Ahí tienes todos los ejercicios resueltos. ¡A matemáticas se ha dicho!

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