Factorization of x⁴ + x²y² + y⁴

What you should know before…

a² – b² = (a + b) (a – b) (1)

a³ + b³ = (a + b) (a² - ab + b²) (2)

a³ – b³ = (a – b) (a² + ab + b²) (3)

a² + 2ab + b² = (a + b)² (4)

The Round-about Tour

Let us begin with the factorization of x⁶ – y⁶ in two ways :

(a) x⁶ – y⁶ = (x²)³ – (y²)³ = (x² – y²)[(x²)² + x²y² +(y²)²], by (3)

= (x + y)(x – y)(x⁴ + x²y² + y⁴), by (1)

(b) x⁶ – y⁶ = (x³)² – (y³)² = (x³ + y³)(x³ – y³), by (1)

= (x + y)(x² – xy + y²)(x - y)(x² +xy +y²), by (2) and (3)

Which of the above factorization is correct?

Of course, (b) is the complete factorization, (a) is not.

Comparing the results in (a) and (b), we can get:

x⁴ + x²y² + y⁴ = (x² + xy + y²)(x² –xy + y²)

Further investigation

x⁴ + x²y² + y⁴ = (x⁴ + 2x²y² + y⁴) - x²y² (note that one term is added and subtracted)

= (x² + y²)² – (xy)², , by (4)

= [(x² + y²) + xy] [(x² + y²) – xy] , by (1)

= (x² + xy + y²)(x² –xy + y²)

Similar way

There are some factorization which use the same technique, here is one example:

x⁴ + 4 = (x⁴ + 4x² + 4) – 4x²

= (x² + 2)² – (2x)²

= (x² + 2 + 2x)(x² + 2 – 2x)

= (x² + 2x + 2)(x² – 2x + 2).