Problem 1

Reduce the following expression to its simplest form:

(9*a2*b2-4*b4)*(a2-b2)-(3*a*b-2*b2)⋅(3*a*[a2+b2]-2*b*[b2+3*a*b-a2])*b

Solution

1. Working on the first term. Noting that

   (9*a2*b2-4*b4)=b2*(9*a2-4*b2)

   and
   9*a2-4*b2=(3*a)2-(2*b)2=(3*a-2*b)*(3*a+2*b)
   Hence, the first term becomes

   b2*(3*a-2*b)*(3*a+2*b)*(a2-b2)

2. Now the second term. Write

   3*a*b-2*b2=b*(3*a-2*b)

   Multiplying by the extra factor b gives

   b2*(3*a-2*b)
   Next, simplify
   T=3*a*(a2+b2)-2*b*(b2+3*a*b-a2)
   Expand to obtain

   3*a*(a2+b2)=3*a3+3*a*b2

   and
   -2*b*(b2+3*a*b-a2)=-2*b3-6*a*b2+2*a2*b

   Combining gives

3. T=3*a3+2*a2*b-3*a*b2-2*b3.

Thus the second term becomes

b2*(3*a-2*b)*(3*a3+2*a2*b-3*a*b2-2*b3)

c) Combine the terms. The expression becomes

E=b2*(3*a-2*b)*[(3*a+2*b)*(a2-b2)-(3*a3+2*a2*b-3*a*b2-2*b3)]

Expand

(3*a+2*b)*(a2-b2)=3*a*(a2-b2)+2*b*(a2-b2)=3*a3-3*a*b2+2*a2*b-2*b3. Subtracting the second group results in

S=3*a3-3*a*b2+2*a2-2*b3-3*a3+2*a2*b-3*a*b2-2*b3=0.

d) Since S=0, it follows that

E=b2*(3*a-2*b)⋅0=0.

Answer: 0