# Ngô Quốc Anh

## September 13, 2009

### An algebraic identity for 9th level

Filed under: Các Bài Tập Nhỏ, Linh Tinh — Ngô Quốc Anh @ 21:16

In this topic I will show you how to prove

$\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}} = 3\sqrt {\sqrt[3]{5} - \sqrt[3]{4}}$

In order to prove that fact, we just do as the following: by using

$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$

we obtain

${\Big( {\sqrt[3]{2} + \underbrace {\sqrt[3]{{20}}}_{\sqrt[3]{{{2^2}}}\sqrt[3]{5}} - \underbrace {\sqrt[3]{{25}}}_{\sqrt[3]{{{5^2}}}}} \Big)^2}=\sqrt[3]{{{2^2}}} + \sqrt[3]{{{2^4}}}\sqrt[3]{{{5^2}}} + \sqrt[3]{{{5^4}}} + 2\left( {\underbrace {\sqrt[3]{2}\sqrt[3]{{{2^2}}}}_2\sqrt[3]{5} - \sqrt[3]{2}\sqrt[3]{{{5^2}}} - \sqrt[3]{{{2^2}}}\underbrace {\sqrt[3]{5}\sqrt[3]{{{5^2}}}}_5} \right).$

Therefore

${\left( {\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}} \right)^2} = \sqrt[3]{{{2^2}}} + 2\sqrt[3]{2}\sqrt[3]{{{5^2}}} + 5\sqrt[3]{5} + 2\left( {2\sqrt[3]{5} - \sqrt[3]{2}\sqrt[3]{{{5^2}}} - 5\sqrt[3]{{{2^2}}}} \right).$

Clearly

$\sqrt[3]{{{2^2}}} + 2\sqrt[3]{2}\sqrt[3]{{{5^2}}} + 5\sqrt[3]{5} + 2\left( {2\sqrt[3]{5} - \sqrt[3]{2}\sqrt[3]{{{5^2}}} - 5\sqrt[3]{{{2^2}}}} \right) = 9\left( {\sqrt[3]{5} - \sqrt[3]{{{2^2}}}} \right).$

Thus

${\left( {\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}} \right)^2} = 9\left( {\sqrt[3]{5} - \sqrt[3]{4}} \right)$

which yields

$\left| {\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}} \right| = 3\sqrt {\sqrt[3]{5} - \sqrt[3]{4}} .$

Finally, by using the fact that $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ we get

${\left( {\sqrt[3]{2} + \sqrt[3]{{20}}} \right)^3} - 25 = \underbrace {\left( {22 + 3\sqrt[3]{{{2^2}}}\sqrt[3]{{20}} + 3\sqrt[3]{2}\sqrt[3]{{{{20}^2}}}} \right) - 25}_{3\left( {\sqrt[3]{{{2^2}}}\sqrt[3]{{20}} + \sqrt[3]{2}\sqrt[3]{{{{20}^2}}} - 1} \right)} > 0$

which implies

${\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}}}>0.$

In other words,

$\sqrt[3]{2} + \sqrt[3]{{20}} - \sqrt[3]{{25}} = 3\sqrt {\sqrt[3]{5} - \sqrt[3]{4}}.$

2. $\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25}=\sqrt[3]{-63-9\sqrt[3]{10}+18\sqrt[3]{100}}$