Question

Find cube root of \sqrt[3]{72-32\sqrt{5}}



Collected in the board: Cube root

Steven Zheng posted 1 week ago

Answer

Let

72+32\sqrt{5} = (a+b\sqrt{5})^3

72-32\sqrt{5} = (a-b\sqrt{5})^3

in which a and b are positive integers

Addition of the two equations gives

(a+b\sqrt{5})^3+(a-b\sqrt{5})^3 = 144

Factoring the LHS using sum of cubes formula.

2a[(a+b\sqrt{5})^2+(a-b\sqrt{5})^2-a^2+5b^2)=144

a(2a^2+2b^2\cdot 5 - a^2+5b^2) = 72

a^3 + 15ab^2 = 72
(1)

Multiplying the two equations gives

(72+32\sqrt{5})(72-32\sqrt{5}) = (a+b\sqrt{5})^3 (a-b\sqrt{5})^3

Simplify the equation using difference of square formula

(a^2 - b^2\cdot 5)^3 = 72^2 - 32^2\cdot 5=64

Taking cube root gives

a^2-5b^2 = 4
(2)

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

\begin{cases} a^3 + 15ab^2 &=72 \\ a^2-5b^2 &=4 \end{cases}

Solving the equation yields

a = 3, b=1

Therefore,

The cube root of

\sqrt[3]{72+32\sqrt{5}}=\sqrt[3]{(3+\sqrt{5})^3}=3+\sqrt{5}

\sqrt[3]{72-32\sqrt{5}}=\sqrt[3]{(3-\sqrt{5})^3}=3-\sqrt{5}


Steven Zheng posted 1 week ago

Scroll to Top