Question
Find cube root of \sqrt[3]{26+15\sqrt{3}}
Find cube root of \sqrt[3]{26+15\sqrt{3}}
Let
26+15\sqrt{3} = (a+b\sqrt{3})^3
26-15\sqrt{3} = (a-b\sqrt{3})^3
in which a and b are positive integers
Addition of the two equations gives
(a+b\sqrt{3})^3+(a-b\sqrt{3})^3 = 52
Factoring the LHS using sum of cubes formula.
2a[(a+b\sqrt{3})^2+(a-b\sqrt{3})^2-a^2+3b^2)=52
a(2a^2+2b^2\cdot 3 - a^2+3b^2) = 26
Multiplying the two equations gives
(26+15\sqrt{3})(26-15\sqrt{3}) = (a+b\sqrt{3})^3 (a-b\sqrt{3})^3
Simplify the equation using difference of square formula
(a^2 - b^2\cdot 3)^3 = 26^2 - 15^2\cdot 3=1
Taking cube root gives
A system of equations is obtained from (1) and (2)
\begin{cases} a^3 + 9ab^2 &=26 \\ a^2-3b^2 &=1 \end{cases}
Solving the equation yields
a = 2, b=1
Therefore,
The cube root of
\sqrt[3]{26+15\sqrt{3}}=\sqrt[3]{(2+\sqrt{3})^3}=2+\sqrt{3}
\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{(2-\sqrt{3})^3}=2-\sqrt{3}