Question

Find cube root of \sqrt[3]{72+32\sqrt{5}}


Collected in the board: Cube root

Steven Zheng posted 6 days ago

Answer

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

a^3 + 15ab^2 = 38
(1)

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^2-5b^2 = -1
(2)

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}

Steven Zheng posted 6 days ago

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