Can you help me solve this Rational Expression?

Question

p^9 - q^9 over p^6 - q^6

Thank you!

Answer 1

p^9-q^9/p^6-q^6
=(p^3-q^3)(p^6+p^3q^3+q^6)/(p^3+q^3)(p^3-q^3)
=(p^6+p^3q^3+q^6)/(p^3+q^3)

Answer 2

No, I cn't help you solve it, because you have no equation to solve.


I'm sure you mean SIMPLIFY instead.

Firstly, don't cancel ANYTHING, because you can only cancel FACTORS in a fraction, i.e. you can only cancel parts of multiplication problems. Since both sides of this fraction are subtraction problems, there is no canceling, at least not YET.

I would recommend factoring the top as a difference of cubes. The formula is

A^3 - B^3 = (A - B)(A^2 + AB + B^2)

Rewrite the top as (p^3)^3 - (q^3)^3. This shows you can let A = p^3 and B = q^3. So the top would factor into
(p^3 - q^3) ( (p^3)^2 + p^3 q^3 + (q^3)^2 ), or
(p^3 - q^3) (p^6 + p^3 q^3 + q^6)

I would factor the bottom as a difference or squares. Yes, it could also be factored as a difference of cubes, but I want to get p^3 and q^3 in the denominator in hopes of getting stuff to match the cubes in the numerator.

The bottom factors into (p^3 - q^3)(p^3 + q^3).

Cancel what you can to get (p^6 + p^3 q^3 + q^6) over (p^3 - q^3). You COULD factor the denominator further (as a sum of cubes), but the numerator won't factor into anything that would cause more canceling.

Answer 3

A = p^9 - q^9 = (p^3)^3 - (q^3)^3 = (p^3 - q^3) (p^6 + q^6 + p^3 * q^3)
B = p^6 - q^6 = (p^3)^2 - (q^3)^2 = (p^3 - q^3) (p^3+q^3)

=> A over B = (p^6 + q^6 + q^3 * p^3) over (p^3 + q^3)