Question

[tex]Se\ considera\ numerele\ :a= \sqrt[3]{41+14\sqrt{109}}+\sqrt[3]{41-14\sqrt{109}} \ si\\
b=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}.\ Sa\ se\ calculeze\ (a-b)^{2006}.[/tex]

Answer 1

$Voi~folosi~identitatea~(x+y)^3=x^3+y^3+3xy(x+y). \\ \\ A= \sqrt[3]{41+14 \sqrt{109}}+ \sqrt[3]{41-14 \sqrt{109}} \\ \\ A^3=82+3 \sqrt[3]{41^2-14 ^2 \cdot 109} \cdot A \\ \\ A^3=82-81A \\ \\ A^3+81A-82=0 \Leftrightarrow (A-1)(A^2+A+82)=0 \Rightarrow A=1. \\ \\ La~fel~se~calculeaza~si~B... \\ \\ B^3=14+3 \sqrt[3]{7^2-5^2 \cdot 2} \cdot B \\ \\ B^3+3B-14=0 \Rightarrow (B-2)(B^2+2B+7)=0 \Rightarrow B=2. \\ \\ .$

$(A-B)^{2006}=1.$