Question

una ayuda para productos notables​

Answer 1

Respuesta:

2

Explicación paso a paso:

$\left(3^n-1\right)\left(9^n+3^n+1\right)=728$

$\left(3^n-1\right)\left(\left(3^2\right)^n+3^n+1\right)=728$

$\left(3^n-1\right)\left(3^{2n}+3^n+1\right)=728$

$\left(3^n-1\right)\left(\left(3^n\right)^2+3^n+1\right)=728$

Re escribir la ecuación con $3^{n}=u$

$\left(u-1\right)\left(\left(u\right)^2+u+1\right)=728$

$\:\left(u-1\right)\left(u^2+u+1\right)=728$

Diferencia de cubos: "$(a-b)(a^{2}+ab+b^{2})=a^{3}-b^{3}$

$u^{3}-1^{3}=728$

$u^{3}-1=728$

$u^{3}=729$

$u=9$

Entonces:

$3^{n}=9\\  \\3^{n}=3^{2}\\  \\n=2$

Answer 2

$(3^{n} -1)(9^{n}+3^{n}+ 1 )= 728\\\\(3^{n} -1)(3^{2n}+3^{n}+ 1 )= 728\\\\27^{n} -1 = 728\\\\27^{n} = 728+1\\\\27^{n} = 729\\\\3^{3n} = 3^{6} \\\\3n = {6}\\\\ n = \frac{6}{3}\\ n = 2$

El valor de n  = 2

Eternamente axllxa