﻿ Find cube root of \sqrt{20+14\sqrt{2}}

#### Question

Find cube root of \sqrt{20+14\sqrt{2}}

Collected in the board: Cube root

Steven Zheng posted 5 months ago

Let

20+14\sqrt{2} = (a+b\sqrt{2})^3

20-14\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 = 40

Factoring the LHS using sum of cubes formula.

2a[(a+b\sqrt{2})^2+(a-b\sqrt{2})^2-a^2+2b^2)=40

a(2a^2+2b^2\cdot 2 - a^2+2b^2) = 20

a^3 + 6ab^2 = 20
(1)

Multiplying the two equations gives

(20+14\sqrt{2})(20-14\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 = 20^2 - 14^2\cdot 2=8

Taking cube root gives

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

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

\begin{cases} a^3 + 6ab^2 &=20 \\ a^2-2b^2 &=2 \end{cases}

Solving the equation yields

a = 2, b=1

Therefore,

The cube root of

\sqrt{20+14\sqrt{2}}=\sqrt{(2+\sqrt{2})^3}=2+\sqrt{2}

\sqrt{20-14\sqrt{2}}=\sqrt{(2-\sqrt{2})^3}=2-\sqrt{2}

Steven Zheng posted 5 months ago

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