#### Question

Find cube root of \sqrt[3]{128-64\sqrt{5}}

Collected in the board: Cube root

Steven Zheng posted 1 week ago

Let

128+64\sqrt{5} = (a+b\sqrt{5})^3

128-64\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 = 256

Factoring the LHS using sum of cubes formula.

2a[(a+b\sqrt{5})^2+(a-b\sqrt{5})^2-a^2+5b^2)=256

a(2a^2+2b^2\cdot 5 - a^2+5b^2) = 128

a^3 + 15ab^2 = 128
(1)

Multiplying the two equations gives

(128+64\sqrt{5})(128-64\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 = 128^2 - 64^2\cdot 5=-4096

Taking cube root gives

a^2-5b^2 = -16
(2)

A system of equations is obtained from (1) and (2)

\begin{cases} a^3 + 15ab^2 &=128 \\ a^2-5b^2 &=-16 \end{cases}

Solving the equation yields

a = 2, b=2

Therefore,

The cube root of

\sqrt[3]{128+64\sqrt{5}}=\sqrt[3]{(2+2\sqrt{5})^3}=2+2\sqrt{5}

\sqrt[3]{128-64\sqrt{5}}=\sqrt[3]{(2-2\sqrt{5})^3}=2-2\sqrt{5}

Steven Zheng posted 1 week ago

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