Question

$Aratati~ca~ \sqrt[3]{16+8 \sqrt{5} } - \sqrt[3]{16-8 \sqrt{5} } =2 \sqrt{5}$

Answer 1

$\displaystyle (a-b)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3-3ab(a-b) \\ \\ Deci~ \boxed{(a-b)^3=a^3-b^3-3ab(a-b)}~. \\ \\ Luam~a= \sqrt[3]{16+ 8 \sqrt{5}}~si~b=\sqrt[3]{16- 8 \sqrt{5}}~si~notam~x=a-b~(adica \\ \\ numarul~din~enunt). \\ \\ Avem~a^3-b^3=(16+8 \sqrt{5})-(16-8 \sqrt{5})=16 \sqrt{5}~si \\ \\ ab= \sqrt[3]{(16+8 \sqrt{5})(16- 8 \sqrt{5})}= \sqrt[3]{256-320} = \sqrt[3]{-64} =-4. \\ \\ Deci~relatia~din~chenar~devine~x^3=16 \sqrt{5}+12x \Leftrightarrow \\ \\ \Leftrightarrow x^3-12x-16 \sqrt{5}=0.$

$x^3-20x+8x-16 \sqrt{5}=0 \\ \\ x(x^2-20)+8(x-2 \sqrt{5})=0. \\ \\x(x-2 \sqrt{5})(x+2 \sqrt{5})+8(x-2 \sqrt{5})=0 \\ \\ (x-2 \sqrt{5})(x^2+2 \sqrt{5}x+8)=0. \\ \\ A~doua~paranteza~nu~poate~fi~nula~(discriminat~negativ). \\ \\ Deci~x=2 \sqrt{5}.$