Problem 9

Find (a-b)6 and (x*y-(x2)/(2*y))6 by the Binomial Theorem.

Solution

a) Expand (a-b)6:

The Binomial Theorem gives

(a-b)6=(∑_k=0^6)(([6],[k])*a(6-k)*(-b)k)

Writing the terms individually:

k=0: ([6],[0])*a6*(-b)0=a6

k=1: ([6],[1])*a5*(-b)1=-6*a5*b

k=2: ([6],[2])*a4*(-b)2=15*a4*b2

k=3: ([6],[3])*a3*(-b)3=-20*a3*b3

k=4: ([6],[4])*a2*(-b)4=15*a2*b4

k=5: ([6],[5])*a*(-b)5=-6*a*b5

k=6: ([6],[6])*(-b)6=b6.

Thus,

⇒(a-b)6=a6-6*a5*b+15*a4*b2-20*a3*b3+15*a2*b4-6*a*b5+b6

b) Expand (x*y-(x2)/(2*y))6:

Write the expression in binomial form by setting

u=x*y and v=(x2)/(2*y)

Then,

(x*y-(x2)/(2*y))6=(∑_k=0^6)(([6],[k])*(x*y)(6-k)⋅(-(x2)/(2*y))k)

Simplify the general term:

(T_k)=([6],[k])*(-1)k(x(2*k))/(2k)*(x*y)(6-k)=([6],[k])*(-1)k(x(2*k)*x(6-k)*y(6-k))/(2k)

=([6],[k])*(-1)k(x(6+k)*y(6-k))/(2k)

Thus, the expansion is

(x*y-(x2)/(2*y))6=(∑_k=0^6)((-1)k(x(6+k)*y(6-k))/(2k))

Summing them up gives us:
x6*y6-3*x7*y4+15/4*x8*y2-5/2*x9+15/(16*y2)*x10-3/(16*y4)*x11+(x12)/(64*y6)

Answer:

a)(a-b)6=a6-6*a5*b+15*a4*b2-20*a3*b3+15*a2*b4-6*a*b5+b6
b)
x6*y6-3*x7*y4+15/4*x8*y2-5/2*x9+15/(16*y2)*x10-3/(16*y4)*x11+(x12)/(64*y6)