Question
Find cube root of \sqrt[3]{7+5\sqrt{2}}
Find cube root of \sqrt[3]{7+5\sqrt{2}}
Let
7+5\sqrt{2} = (a+b\sqrt{2})^3
7-5\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 = 14
Factoring the LHS using sum of cubes formula.
2a[(a+b\sqrt{2})^2+(a-b\sqrt{2})^2-a^2+2b^2)=14
a(2a^2+2b^2\cdot 2 - a^2+2b^2) = 7
Multiplying the two equations gives
(7+5\sqrt{2})(7-5\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 = 7^2 - 5^2\cdot 2=-1
Taking cube root gives
A system of equations is obtained from (1) and (2)
\begin{cases} a^3 + 6ab^2 &=7 \\ a^2-2b^2 &=-1 \end{cases}
Solving the equation yields
a = 1, b=1
Therefore,
The cube root of
\sqrt[3]{7+5\sqrt{2}}=\sqrt[3]{(1+\sqrt{2})^3}=1+\sqrt{2}
\sqrt[3]{7-5\sqrt{2}}=\sqrt[3]{(1-\sqrt{2})^3}=1-\sqrt{2}