Expand Using the Binomial Theorem (x-y)^11

${(x - y)}^{11}$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{11} \frac{11!}{(11 - k)! k!} \cdot {(x)}^{11 - k} \cdot {(- y)}^{k}$

Step 2

Expand the summation.

$\frac{11!}{(11 - 0)! 0!} {(x)}^{11 - 0} \cdot {(- y)}^{0} + \frac{11!}{(11 - 1)! 1!} {(x)}^{11 - 1} \cdot {(- y)}^{1} + \dots + \frac{11!}{(11 - 11)! 11!} {(x)}^{11 - 11} \cdot {(- y)}^{11}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{11} \cdot {(- y)}^{0} + 11 \cdot {(x)}^{10} \cdot {(- y)}^{1} + \dots + 1 \cdot {(x)}^{0} \cdot {(- y)}^{11}$

Step 4

Simplify the polynomial result.

$x^{11} - 11 x^{10} y + 55 x^{9} y^{2} - 165 x^{8} y^{3} + 330 x^{7} y^{4} - 462 x^{6} y^{5} + 462 x^{5} y^{6} - 330 x^{4} y^{7} + 165 x^{3} y^{8} - 55 x^{2} y^{9} + 11 x y^{10} - y^{11}$

Expand Using the Binomial Theorem (x-y)^11

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