﻿ Find cube root of \sqrt[3]{26+15\sqrt{3}}

#### Question

Find cube root of \sqrt[3]{26+15\sqrt{3}}

Collected in the board: Cube root

Steven Zheng posted 5 months ago

Let

26+15\sqrt{3} = (a+b\sqrt{3})^3

26-15\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 = 52

Factoring the LHS using sum of cubes formula.

2a[(a+b\sqrt{3})^2+(a-b\sqrt{3})^2-a^2+3b^2)=52

a(2a^2+2b^2\cdot 3 - a^2+3b^2) = 26

a^3 + 9ab^2 = 26
(1)

Multiplying the two equations gives

(26+15\sqrt{3})(26-15\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 = 26^2 - 15^2\cdot 3=1

Taking cube root gives

a^2-3b^2 = 1
(2)

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

\begin{cases} a^3 + 9ab^2 &=26 \\ a^2-3b^2 &=1 \end{cases}

Solving the equation yields

a = 2, b=1

Therefore,

The cube root of

\sqrt[3]{26+15\sqrt{3}}=\sqrt[3]{(2+\sqrt{3})^3}=2+\sqrt{3}

\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{(2-\sqrt{3})^3}=2-\sqrt{3}

Steven Zheng posted 5 months ago

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