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.
Dan

Latest posts by Dan (see all)

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *