Question

$Calculati: \\ \\ \sqrt[3]{(10+ \sqrt{68}) } ~* \sqrt[3]{(10- \sqrt{68}) } : \sqrt[6]{16}$

Answer 1

Trebuie sa aplici formula de calcul prescurtat: (a-b)(a+b)=a²-b²

$\:\sqrt[3]{\left(10+\sqrt{68}\right)}\cdot \sqrt[3]{\left(10-\sqrt{68}\right)}:\:\sqrt[6]{16} =$
$= \:\sqrt[3]{\left(10+\sqrt{68}\right)\left(10-\sqrt{68}\right)}:\:\sqrt[6]{16} =$ 
$= \:\sqrt[3]{100 - 68}:\:\sqrt[6]{16} =$
$= \:\sqrt[3]{32}:\:\sqrt[6]{16} = \sqrt[3]{8\cdot4}:\sqrt[6]{2^4}=2\sqrt[3]{4}:\sqrt[6]{2^4} =$
$=2\cdot 2^{\frac{2}{3}}:2^{\frac{2}{3}}=$
$= 2$

Answer 2

$[tex]\it \sqrt[6]{16} = 16^{\frac{1}{6}} = (4^2) ^\frac{1}{6} = 4^{2\cdot\frac{1}{6}} = 4^{\frac{1}{3}} = \sqrt[3]{\it 4}$ [/tex]

Acum, noi putem folosi un singur radical,  peste toată expresia:

$\it \sqrt[3]{(\it 10+\sqrt{68})(10-\sqrt{68}):4} = \sqrt[3]{(\it 100-68):4} =\sqrt[3]{\it 32:4} =\sqrt[3]{\it8} =2$