﻿ Find cube root of \sqrt[3]{7+5\sqrt{2}}

#### Question

Find cube root of \sqrt[3]{7+5\sqrt{2}}

Collected in the board: Cube root

Steven Zheng posted 1 week ago

Let

7+5\sqrt{2} = (a+b\sqrt{2})^3

7-5\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 = 14

Factoring the LHS using sum of cubes formula.

2a[(a+b\sqrt{2})^2+(a-b\sqrt{2})^2-a^2+2b^2)=14

a(2a^2+2b^2\cdot 2 - a^2+2b^2) = 7

a^3 + 6ab^2 = 7
(1)

Multiplying the two equations gives

(7+5\sqrt{2})(7-5\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 = 7^2 - 5^2\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 &=7 \\ a^2-2b^2 &=-1 \end{cases}

Solving the equation yields

a = 1, b=1

Therefore,

The cube root of

\sqrt[3]{7+5\sqrt{2}}=\sqrt[3]{(1+\sqrt{2})^3}=1+\sqrt{2}

\sqrt[3]{7-5\sqrt{2}}=\sqrt[3]{(1-\sqrt{2})^3}=1-\sqrt{2}

Steven Zheng posted 1 week ago

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