Question

pre calculo alguém me ajuda

Answer 1

$\boxed{(A-B)^3 =A^3-3*A^2*B + 3*A*B^2-B^3}$

temos
$( \sqrt[6]{7} - \sqrt[9]{3} )^3= (7^{ \frac{1}{6}} - 3^{ \frac{1}{9} } })^3$

temos
$(7^{ \frac{1}{6}})^3 - 3* (7^{ \frac{1}{6}})^2*3^{ \frac{1}{9} }+3*7^{ \frac{1}{6}}*(3^{ \frac{1}{9} })^2- (3^{ \frac{1}{9} })^3\\\\ =(7^{ \frac{3}{6}}) - (7^{ \frac{2}{6}})*3^{1+ \frac{1}{9} }+7^{ \frac{1}{6}}*(3^{1+ \frac{2}{9} })- (3^{ \frac{3}{9} })\\\\ =(7^{ \frac{1}{2}}) - (7^{ \frac{1}{3}})*3^{ \frac{10}{9} }+7^{ \frac{1}{6}}*(3^{ \frac{11}{9} })- (3^{ \frac{1}{3} })\\\\= \sqrt{7}- \sqrt[3]{7}* \sqrt[9]{3^{10}}+ \sqrt[6]{7} * \sqrt[9]{3^{11}} - \sqrt[3]{3}$