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.

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