Question

Find cube root of \sqrt[3]{10+6\sqrt{3}}


Collected in the board: Cube root

Steven Zheng posted 1 week ago

Answer

Let

10+6\sqrt{3} = (a+b\sqrt{3})^3

10-6\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 = 20

Factoring the LHS using sum of cubes formula.

2a[(a+b\sqrt{3})^2+(a-b\sqrt{3})^2-a^2+3b^2)=20

a(2a^2+2b^2\cdot 3 - a^2+3b^2) = 10

a^3 + 9ab^2 = 10
(1)

Multiplying the two equations gives

(10+6\sqrt{3})(10-6\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 = 10^2 - 6^2\cdot 3=-8

Taking cube root gives

a^2-3b^2 = -2
(2)

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

\begin{cases} a^3 + 9ab^2 &=10 \\ a^2-3b^2 &=-2 \end{cases}

Solving the equation yields

a = 1, b=1

Therefore,

The cube root of

\sqrt[3]{10+6\sqrt{3}}=\sqrt[3]{(1+\sqrt{3})^3}=1+\sqrt{3}

\sqrt[3]{10-6\sqrt{3}}=\sqrt[3]{(1-\sqrt{3})^3}=1-\sqrt{3}

Steven Zheng posted 1 week ago

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