Question

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Answer 1

Respuesta:

$\frac{x^3-1}{7x^4-7}=\frac{x^2+x+1}{7\left(x^2+1\right)\left(x+1\right)}$

Explicación paso a paso:

Factorizar $x^3-1=\left(x-1\right)\left(x^2+x+1\right)$

Factorizar $7x^4-7=7\left(x^4-1\right)$

Factorizar $x^4-1=\left(x^2\right)^2-1^2$

Factorizar $\left(x^2\right)^2-1^2=\left(x^2+1\right)\left(x^2-1\right)$

Factorizar $x^2-1=\left(x+1\right)\left(x-1\right)$

Factorizar $=\frac{\left(x-1\right)\left(x^2+x+1\right)}{7\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}$

Eliminar los términos comunes x-1

$=\frac{x^2+x+1}{7\left(x^2+1\right)\left(x+1\right)}$