Question

Find cube root of \sqrt[3]{99+70\sqrt{2}}


Collected in the board: Cube root

Steven Zheng posted 6 days ago

Answer

Let

99+70\sqrt{2} = (a+b\sqrt{2})^3

99-70\sqrt{2} = (a-b\sqrt{2})^3

in which a and b are positive integers

Addition of the two equations gives

(a+b\sqrt{2})^3+(a-b\sqrt{2})^3 = 198

Factoring the LHS using sum of cubes formula.

2a[(a+b\sqrt{2})^2+(a-b\sqrt{2})^2-a^2+2b^2)=198

a(2a^2+2b^2\cdot 2 - a^2+2b^2) = 99

a^3 + 6ab^2 = 99
(1)

Multiplying the two equations gives

(99+70\sqrt{2})(99-70\sqrt{2}) = (a+b\sqrt{2})^3 (a-b\sqrt{2})^3

Simplify the equation using difference of square formula

(a^2 - b^2\cdot 2)^3 = 99^2 - 70^2\cdot 2=1

Taking cube root gives

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

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

\begin{cases} a^3 + 6ab^2 &=99 \\ a^2-2b^2 &=1 \end{cases}

Solving the equation yields

a = 3, b=2

Therefore,

The cube root of

\sqrt[3]{99+70\sqrt{2}}=\sqrt[3]{(3+2\sqrt{2})^3}=3+2\sqrt{2}

\sqrt[3]{99-70\sqrt{2}}=\sqrt[3]{(3-2\sqrt{2})^3}=3-2\sqrt{2}

Steven Zheng posted 6 days ago

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