Question

Simplifique as expressões, supondo ab≠0:

(a^-3 - b^-3) : (a^-1 - b^-1)

Answer 1

$\\a^{-3} - b^{-3}=(\frac{1}{a}\big)^3-(\frac{1}{b}\big)^3 =(\frac{1}{a}-\frac{1}{b}\big)\cdot \big[\big( \frac{1}{a}\big)^2 + \frac{1}{ab} + \big(\frac{1}{b}\big)^2\big]\\\\-------------------------\\a^{-1} - b^{-1}=\frac{1}{a}-\frac{1}{b}\big\\\\-------------------------\\(a^{-3} - b^{-3}) : (a^{-1} - b^{-1})=\frac{a^{-3} - b^{-3}}{a^{-1} - b^{-1}}$

$\\-------------------------\\(a^{-3} - b^{-3}) : (a^{-1} - b^{-1})=\frac{a^{-3} - b^{-3}}{a^{-1} - b^{-1}} =\frac{\big(\frac{1}{a}-\frac{1}{b}\big)\cdot \big[\big( \frac{1}{a}\big)^2 + \frac{1}{ab} + \big(\frac{1}{b}\big)^2\big]}{\frac{1}{a}-\frac{1}{b}} \\\\=\big( \frac{1}{a}\big)^2 + \frac{1}{ab} + \big(\frac{1}{b}\big)^2$

Também pode ser:

$=\frac{1}{a^2} + \frac{1}{ab} + \frac{1}{b^2}$

ou ainda:

$=\frac{1}{a^2} + \frac{1}{ab} + \frac{1}{b^2}\\\\=a^{-2} +a^{-1}b^{-1} + b^{-2}$