Question

factoriza las siguientes diferencias de cuadrados perfectos​

Answer 1

Respuesta:

[tex]i {}^{16} - z {}^{4} = (i {}^{8} + z {}^{2})( (i {}^{8} - z {}^{2} )[/tex]

[tex]400p {}^{6} - 256q {}^{4} = (20p {}^{3} + 16q {}^{2} )(20p {}^{3} - 16q {}^{2})[/tex]

Answer 2

Verified answer

[tex]\bold{=\left(r^8\right)^2-\left(z^2\right)^2}[/tex]

[tex]\mathrm{Aplicar\:la\:siguiente\:regla\:para\:binomios\:al\:cuadrado:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)[/tex]

[tex]\bold{=\left(r^8+z^2\right)\left(r^8-z^2\right)}[/tex]

[tex]\mathrm{Factorizar}\:r^{8}-z^2[/tex]

[tex]\boxed{\bold{=\left(r^8+z^2\right)\left(r^4+z\right)\left(r^4-z\right)}}[/tex]

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[tex]\bold{=16\left(25p^6-16q^4\right)}[/tex]

[tex]\mathrm{Factorizar}\:25p^6-16q^4[/tex]

[tex]\boxed{\bold{=16\left(5p^3+4q^2\right)\left(5p^3-4q^2\right)}}[/tex]