Question

simplify (p+q)(p^2-q^2)+(p-q)(p^2+q^3)

Answer 1

-p3q + p3 - 2p2q2 + q
—————————————————————
p2
Step by step solution :
Step 1 :
Equation at the end of step 1 :
q
((p + ————) - pq) - 2q2
(p2)
Step 2 :
q
Simplify ——
p2
Equation at the end of step 2 :
q
((p + ——) - pq) - 2q2
p2
Step 3 :
Rewriting the whole as an Equivalent Fraction :
3.1 Adding a fraction to a whole

Rewrite the whole as a fraction using p2 as the denominator :

p p • p2
p = — = ——————
1 p2
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :
3.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

p • p2 + q p3 + q
—————————— = ——————
p2 p2
Equation at the end of step 3 :
(p3 + q)
(———————— - pq) - 2q2
p2
Step 4 :
Rewriting the whole as an Equivalent Fraction :
4.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using p2 as the denominator :

pq pq • p2
pq = —— = ———————
1 p2
Trying to factor as a Sum of Cubes :
4.2 Factoring: p3 + q

Theory : A sum of two perfect cubes, a3 + b3 can be factored into :
(a+b) • (a2-ab+b2)
Proof : (a+b) • (a2-ab+b2) =
a3-a2b+ab2+ba2-b2a+b3 =
a3+(a2b-ba2)+(ab2-b2a)+b3=
a3+0+0+b3=
a3+b3

Check : p3 is the cube of p1

Check : q 1 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes


Adding fractions that have a common denominator :
4.3 Adding up the two equivalent fractions

(p3+q) - (pq • p2) -p3q + p3 + q
—————————————————— = —————————————
p2 p2
Equation at the end of step 4 :
(-p3q + p3 + q)
——————————————— - 2q2
p2
Step 5 :
Rewriting the whole as an Equivalent Fraction :
5.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using p2 as the denominator :

2q2 2q2 • p2
2q2 = ——— = ————————
1 p2
Trying to factor a multi variable polynomial :
5.2 Factoring -p3q + p3 + q

Try to factor this multi-variable trinomial using trial and error

Factorization fails

Adding fractions that have a common denominator :
5.3 Adding up the two equivalent fractions

(-p3q+p3+q) - (2q2 • p2) -p3q + p3 - 2p2q2 + q
———————————————————————— = —————————————————————
p2 p2
Checking for a perfect cube :
5.4 -p3q + p3 - 2p2q2 + q is not a perfect cube

Final result :
-p3q + p3 - 2p2q2 + q
—————————————————————
p2