Question

não entendi essas partes


(FEI-SP-95) Simplificando a expressão representada a seguir, obtemos (a²b+ab²)*(1/a³-1/b³)/(1/a²-1/b²). ​

Answer 1

$\large\boxed{\begin{array}{l}\rm lembre-se\,que\\\rm a^2b+ab^2=ab(a+b)\\\rm\dfrac{1}{a^3}-\dfrac{1}{b^3}=\dfrac{b^3-a^3}{a^3\cdot b^3}=\dfrac{(b-a)(b^2+ab+a^2)}{a^3\cdot b^3}\\\\\rm\dfrac{1}{a^2}-\dfrac{1}{b^2}=\dfrac{b^2-a^2}{a^2\cdot b^2}=\dfrac{(b-a)(b+a)}{a^2\cdot b^2}\end{array}}$

$\large\boxed{\begin{array}{l}\rm(a^2b+ab^2)\cdot\dfrac{\bigg(\dfrac{1}{a^3}-\dfrac{1}{b^3}\bigg)}{\bigg(\dfrac{1}{a^2}-\dfrac{1}{b^2}\bigg)}\\\\\rm substituindo\,cada\,termo\,pela\,sua\,forma\\\rm fatorada\,temos:\\\rm ab\cdot(a+b)\cdot\dfrac{\dfrac{(b-a)(b^2+ab+a^2)}{a^3\cdot b^3}}{\dfrac{(b-a)(b+a)}{a^2\cdot b^2}}\\\\\sf Resolvendo\,a\,divis\tilde ao\,de\,frac_{\!\!,}\tilde ao\,temos:\\\rm ab\cdot(a+b)\cdot\dfrac{(b-a)(b^2+ab+a^2)}{a^3\cdot b^3}\cdot\dfrac{a^2\cdot b^2}{(b-a)(b+a)}\end{array}}$

$\large\boxed{\begin{array}{l}\sf Perceba\,que\,podemos\,simplificar\\\sf as\,express\tilde oes~a+b~com~b+a~e~b-a~com~b-a.\\\rm ab\cdot\diagdown\!\!\!\!\!(a+\diagdown\!\!\!\!b)\dfrac{\diagdown\!\!\!\!\!\!(b-\diagdown\!\!\!\!a)(b^2+ab+a^2)}{a^3\cdot b^3}\cdot\dfrac{a^2\cdot b^2}{\diagdown\!\!\!\!\!\!(b-\diagdown\!\!\!\!a)\diagdown\!\!\!\!\!\!(b+\diagdown\!\!\!\!a)}\\\sf vamos\,multiplicar\,ab\,por\,a^2\cdot b^2:\\\rm\dfrac{(b^2+ab+a^2)\cdot a^3\cdot b^3}{a^3\cdot b^3}\end{array}}$

$\large\boxed{\begin{array}{l}\rm simplificando\,a^3\cdot b^3\,com\,a^3\cdot b^3\,temos:\\\rm\dfrac{(b^2+ab+a^2)\cdot\diagup\!\!\!\!a^3\cdot\diagup\!\!\!\!b^3}{\diagup\!\!\!\!a^3\cdot \diagup\!\!\!\!b^3}=\huge\boxed{\boxed{\boxed{\boxed{\rm a^2+ab+b^2}}}}\end{array}}$