When is it to use simpler numbers to solve a problem

Work Shown:

p-q = -7

(p-q)^2 = (-7)^2

p^2-2pq+q^2 = 49

p^2-2(-10)+q^2 = 49

p^2+20+q^2 = 49

p^2+q^2 = 49-20

p^2+q^2 = 29

By the difference of cubes formula, we can say,

p^3 - q^3 = (p-q)*(p^2+pq+q^2)

p^3 - q^3 = (p-q)*(p^2+q^2+pq)

p^3 - q^3 = (-7)*(29-10)

p^3 - q^3 = -133

Here's another way to solve:

p-q = -7

p = q-7

pq = -10

(q-7)*q = -10

q^2-7q = -10

q^2-7q+10 = 0

(q-5)(q-2) = 0

q-5 = 0 or q-2 = 0

q = 5 or q = 2

If q = 5, then p = q-7 = 5-7 = -2

If q = 2, then p = q-7 = 2-7 = -5

We can see that no matter what we pick for q, we'll get p*q = -10

Also, you should find that p^3-q^3 is equal to -133 for either case as well.