Question

resolver (4y4+5)3 ayudenmen porfa :)​

Answer 1

[tex]La\:respuesta\:es[/tex] ⇒[tex]64y^{12}+240y^8+300y^4+125[/tex]

[tex]Veamos\:los\:pasos[/tex]

[tex]\left(4y^4+5\right)^3[/tex]

[tex]\mathrm{Comos\:sabemos,\:aplicamos\:la\:formula\:del\:binomio\:al\:cubo}:\quad \left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3[/tex]

[tex]\left(4y^4+5\right)^3=\left(4y^4\right)^3+3\left(4y^4\right)^2\cdot \:5+3\cdot \:4y^4\cdot \:5^2+5^3[/tex]

[tex]=\left(4y^4\right)^3+3\left(4y^4\right)^2\cdot \:5+3\cdot \:4y^4\cdot \:5^2+5^3[/tex]

[tex]Simplificamos \left(4y^4\right)^3+3\left(4y^4\right)^2\cdot \:5+3\cdot \:4y^4\cdot \:5^2+5^3: 64y^{12} +240y^{8} +300y^{4} +125[/tex]

[tex]=64y^{12}+240y^8+300y^4+125[/tex]

[tex]Y\:la\:otra\:que\:me\:dijiste[/tex]

[tex]La\:respuesta\:es[/tex] ⇒[tex]125x^6-225x^4+135x^2-27[/tex]

[tex]Veamos\:los\:pasos[/tex]

[tex]\left(5x^2-3\right)^3[/tex]

[tex]\left(5x^2-3\right)^3=\left(5x^2\right)^3-3\left(5x^2\right)^2\cdot \:3+3\cdot \:5x^2\cdot \:3^2-3^3[/tex]

[tex]=\left(5x^2\right)^3-3\left(5x^2\right)^2\cdot \:3+3\cdot \:5x^2\cdot \:3^2-3^3[/tex]

[tex]=125x^6-225x^4+135x^2-27[/tex]

[tex]Buena\:suerte\:con\:tus\:tareas\::ProfeAndresFelipe :)[/tex]