# Tricky Algebraic Identity

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

**Problem:**

If , find the numerical value of .

**Solution 1:**

Using the following identities:

Since

by substitution,

Also

Let and

Derived from above

Use any method to solve for u:

By factoring,

and the other values of u’s are imaginary.

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

### Dan

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