# Tricky Algebraic Identity

Today’s featured problem came from the Mathematician of Technological Institute of the Philippines – Quezon City, Joven Marzan Baltazar. He placed 2nd in the inter-department quiz bee at the same school. The problem is about tricky algebraic identity. Let’s take a look.

Problem:

If $x^3+\displaystyle\frac{1}{x^3}=2\sqrt{5}$, find the numerical value of $x^2+\displaystyle\frac{1}{x^2}$.

Solution 1:

Using the following identities:

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

$(x^3+\displaystyle\frac{1}{x^3})^2=(x^3)^2+2(x^3)( \displaystyle\frac{1}{x^3})+(\displaystyle\frac{1}{x^3})^2$ $(x^3+\displaystyle\frac{1}{x^3})^2=x^6+2+\displaystyle\frac{1}{x^6}$

Since $x^3+\displaystyle\frac{1}{x^3}=2\sqrt{5}$

by substitution,

$(2\sqrt{5})^2=x^6+2+\displaystyle\frac{1}{x^6}$ $20=x^6+2+\displaystyle\frac{1}{x^6}$ $x^6+\displaystyle\frac{1}{x^6}=18$

Also

$(x^2+\displaystyle\frac{1}{x^2})^3=(x^2)^3+3(x^2)^2(\displaystyle\frac{1}{x^2})+3(x^2)( \displaystyle\frac{1}{x^2})^2+(\displaystyle\frac{1}{x^2})^3$ $(x^2+\displaystyle\frac{1}{x^2})^3=x^6+3x^2+\displaystyle\frac{3}{x^2}+\displaystyle\frac{1}{x^6}$ $(x^2+\displaystyle\frac{1}{x^2})^3=x^6+\displaystyle\frac{1}{x^6}+3(x^2+\displaystyle\frac{1}{x^2})$

Let $u=x^2+\displaystyle\frac{1}{x^2}$ and $x^6+\displaystyle\frac{1}{x^6}=18$

Derived from above

$(u)^3=18 +3(u)$ $u^3-3u-18=0$

Use any method to solve for u:

By factoring,

$(u-3)(u^2+3u+18)=0$

$u=3$ and the other values of u’s are imaginary.

$u=x^2+\displaystyle\frac{1}{x^2}=3$

There are a lot of other ways to solve this problem. If you have some let us know.

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

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