[tex](2- \sqrt[3]{6})(4+2 \sqrt[3]{6} + \sqrt[3]{36}) =[/tex] помогите ,пожалуйста!!. Created by nklad44. algebra-ru

tex] помогите ,пожалуйста!!...

RussianMastermind

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= 8 + 4*6^1/3 + 2*6^2/3 - 4*6^1/3 - 2*6^2/3 - 6 = 2

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Luluput

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$(2- \sqrt[3]{6} )(4+2 \sqrt[3]{6}+ \sqrt[3]{36})=(2- \sqrt[3]{6} )(2^2+2 \sqrt[3]{6}+ (\sqrt[3]{6})^2)=2^3-( \sqrt[3]{6} )^3=$ $=8-6=2$

P. S.

$a^3-b^3=(a-b)(a^2+ab+b^2)$

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