Problem 9

Find ${(a-b)}^6$ and ${(xy-\frac{x^2}{2y})}^6$ by the Binomial Theorem.

Solution

a) Expand ${(a-b)}^6$:

The Binomial Theorem gives

${(a-b)}^6={\sum }_{k=0}^6(\left(\begin{array}{c}6\\ k\end{array}\right)a^{6-k}{(-b)}^k)$

Writing the terms individually:

$k=0:$ $\left(\begin{array}{c}6\\ 0\end{array}\right)a^6{(-b)}^0=a^6$

$k=1:$ $\left(\begin{array}{c}6\\ 1\end{array}\right)a^5{(-b)}^1=-6a^{5}b$

$k=2:$ $\left(\begin{array}{c}6\\ 2\end{array}\right)a^4{(-b)}^2=15a^4{b}^2$

$k=3:$ $\left(\begin{array}{c}6\\ 3\end{array}\right)a^3{(-b)}^3=-20a^3{b}^3$

$k=4:$ $\left(\begin{array}{c}6\\ 4\end{array}\right)a^2{(-b)}^4=15a^2{b}^4$

$k=5:$ $\left(\begin{array}{c}6\\ 5\end{array}\right)a{(-b)}^5=-6a{b}^5$

$k=6:$ $\left(\begin{array}{c}6\\ 6\end{array}\right){(-b)}^6=b^{6.}$

Thus,

$\Rightarrow {(a-b)}^6=a^6-6a^{5}b+15a^4{b}^2-20a^3{b}^3+15a^2{b}^4-6a{b}^5+b^6$

b) Expand ${(xy-\frac{x^2}{2y})}^6$:

Write the expression in binomial form by setting

$u=xy$ and $v=\frac{x^2}{2y}$

Then,

${(xy-\frac{x^2}{2y})}^6={\sum }_{k=0}^6(\left(\begin{array}{c}6\\ k\end{array}\right){(xy)}^{6-k}\cdot {(-\frac{x^2}{2y})}^k)$

Simplify the general term:

$T_k=\left(\begin{array}{c}6\\ k\end{array}\right){(-1)}^k\frac{x^{2k}}{2^k}{(xy)}^{6-k}=\left(\begin{array}{c}6\\ k\end{array}\right){(-1)}^k\frac{x^{2k}{x}^{6-k}{y}^{6-k}}{2^k}$

$=\left(\begin{array}{c}6\\ k\end{array}\right){(-1)}^k\frac{x^{6+k}{y}^{6-k}}{2^k}$

Thus, the expansion is

${(xy-\frac{x^2}{2y})}^6={\sum }_{k=0}^6({(-1)}^k\frac{x^{6+k}{y}^{6-k}}{2^k})$

Summing them up gives us:
$x^6{y}^6-3x^7{y}^4+\frac{15}{4}{x}^8{y}^2-\frac{5}{2}{x}^9+\frac{15}{16y^2}{x}^{10}-\frac{3}{16y^4}{x}^{11}+\frac{x^{12}}{64y^6}$

Answer:

a)${(a-b)}^6=a^6-6a^{5}b+15a^4{b}^2-20a^3{b}^3+15a^2{b}^4-6a{b}^5+b^6$
b)
$x^6{y}^6-3x^7{y}^4+\frac{15}{4}{x}^8{y}^2-\frac{5}{2}{x}^9+\frac{15}{16y^2}{x}^{10}-\frac{3}{16y^4}{x}^{11}+\frac{x^{12}}{64y^6}$