﻿ Simplify \sqrt[3]{2+\sqrt{5} }

#### Question

Simplify \sqrt[3]{2+\sqrt{5} }

Collected in the board: Cube root

Steven Zheng posted 2 years ago

Steven Zheng posted 2 years ago

#### This answer is set private or in draft.

\sqrt[3]{2+\sqrt{5} }

=\sqrt[3]{\dfrac{8(2+\sqrt{5})}{8} }

=\sqrt[3]{\dfrac{16+8\sqrt{5} }{8}}
(1)

Now Let the numerator part

16+8\sqrt{5}=(x+y\sqrt{5})^3
(2)

Using the cube of binomial formula

(a+b)^3=a^3+3a^2b + 3ab^2 +b^3

(x+y\sqrt{5})^3 = x^3+3x^2y\sqrt{5}+3x(y\sqrt{5} )^2 +(y\sqrt{5} )^3

= x^3+3x^2y\sqrt{5}+15xy^2 +5y^2\sqrt{5}

=(x^3+15xy^2)+(3x^2y\sqrt{5}+5y^2\sqrt{5})

=(x^3+15xy^2)+(3x^2y+5y^2)\sqrt{5}
(3)

Comparing the equations (2) and (3) gives the following equation system

\begin{cases} x^3+15xy^2 &=16 \\ 3x^2y+5y^2 &=8 \end{cases}

Since x,y are integers, it's easy to solve the system

x=1, y=1

Substituting to (2), the expression (1) is transformed to

\sqrt[3]{\dfrac{16+8\sqrt{5} }{8}}

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

=\dfrac{1+\sqrt{5} }{2}

Steven Zheng posted 2 years ago

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