Question
Find cube root of \sqrt[3]{25+22\sqrt{2}}
Find cube root of \sqrt[3]{25+22\sqrt{2}}
Let
25+22\sqrt{2} = (a+b\sqrt{2})^3
25-22\sqrt{2} = (a-b\sqrt{2})^3
in which a and b are positive integers
Addition of the two equations gives
(a+b\sqrt{2})^3+(a-b\sqrt{2})^3 = 50
Factoring the LHS using sum of cubes formula.
2a[(a+b\sqrt{2})^2+(a-b\sqrt{2})^2-a^2+2b^2)=50
a(2a^2+2b^2\cdot 2 - a^2+2b^2) = 25
Multiplying the two equations gives
(25+22\sqrt{2})(25-22\sqrt{2}) = (a+b\sqrt{2})^3 (a-b\sqrt{2})^3
Simplify the equation using difference of square formula
(a^2 - b^2\cdot 2)^3 = 25^2 - 22^2\cdot 2=-343
Taking cube root gives
A system of equations is obtained from (1) and (2)
\begin{cases} a^3 + 6ab^2 &=25 \\ a^2-2b^2 &=-7 \end{cases}
Solving the equation yields
a = 1, b=2
Therefore,
The cube root of
\sqrt[3]{25+22\sqrt{2}}=\sqrt[3]{(1+2\sqrt{2})^3}=1+2\sqrt{2}
\sqrt[3]{25-22\sqrt{2}}=\sqrt[3]{(1-2\sqrt{2})^3}=1-2\sqrt{2}