Question
Find cube root of \sqrt[3]{72+32\sqrt{5}}
Find cube root of \sqrt[3]{72+32\sqrt{5}}
Let
38+17\sqrt{5} = (a+b\sqrt{5})^3
38-17\sqrt{5} = (a-b\sqrt{5})^3
in which a and b are positive integers
Addition of the two equations gives
(a+b\sqrt{5})^3+(a-b\sqrt{5})^3 = 76
Factoring the LHS using sum of cubes formula.
2a[(a+b\sqrt{5})^2+(a-b\sqrt{5})^2-a^2+5b^2)=76
a(2a^2+2b^2\cdot 5 - a^2+5b^2) = 38
Multiplying the two equations gives
(38+17\sqrt{5})(38-17\sqrt{5}) = (a+b\sqrt{5})^3 (a-b\sqrt{5})^3
Simplify the equation using difference of square formula
(a^2 - b^2\cdot 5)^3 = 38^2 - 17^2\cdot 5=-1
Taking cube root gives
A system of equations is obtained from (1) and (2)
\begin{cases} a^3 + 15ab^2 &=38 \\ a^2-5b^2 &=-1 \end{cases}
Solving the equation yields
a = 2, b=1
Therefore,
The cube root of
\sqrt[3]{38+17\sqrt{5}}=\sqrt[3]{(2+\sqrt{5})^3}=2+\sqrt{5}
\sqrt[3]{38-17\sqrt{5}}=\sqrt[3]{(2-\sqrt{5})^3}=2-\sqrt{5}