Question
Find cube root of \sqrt[3]{72-32\sqrt{5}}
Find cube root of \sqrt[3]{72-32\sqrt{5}}
Let
72+32\sqrt{5} = (a+b\sqrt{5})^3
72-32\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 = 144
Factoring the LHS using sum of cubes formula.
2a[(a+b\sqrt{5})^2+(a-b\sqrt{5})^2-a^2+5b^2)=144
a(2a^2+2b^2\cdot 5 - a^2+5b^2) = 72
Multiplying the two equations gives
(72+32\sqrt{5})(72-32\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 = 72^2 - 32^2\cdot 5=64
Taking cube root gives
A system of equations is obtained from (1) and (2)
\begin{cases} a^3 + 15ab^2 &=72 \\ a^2-5b^2 &=4 \end{cases}
Solving the equation yields
a = 3, b=1
Therefore,
The cube root of
\sqrt[3]{72+32\sqrt{5}}=\sqrt[3]{(3+\sqrt{5})^3}=3+\sqrt{5}
\sqrt[3]{72-32\sqrt{5}}=\sqrt[3]{(3-\sqrt{5})^3}=3-\sqrt{5}