Expand Using the Binomial Theorem ( square root of a+ square root of b)^12

${(\sqrt{a} + \sqrt{b})}^{12}$

Step 1

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

$\sum_{k = 0}^{12} \frac{12!}{(12 - k)! k!} \cdot {(\sqrt{a})}^{12 - k} \cdot {(\sqrt{b})}^{k}$

Step 2

Expand the summation.

$\frac{12!}{(12 - 0)! 0!} \cdot {(\sqrt{a})}^{12 - 0} \cdot {(\sqrt{b})}^{0} + \frac{12!}{(12 - 1)! 1!} \cdot {(\sqrt{a})}^{12 - 1} \cdot {(\sqrt{b})}^{1} + \dots + \frac{12!}{(12 - 12)! 12!} \cdot {(\sqrt{a})}^{12 - 12} \cdot {(\sqrt{b})}^{12}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(\sqrt{a})}^{12} \cdot {(\sqrt{b})}^{0} + 12 \cdot {(\sqrt{a})}^{11} \cdot {(\sqrt{b})}^{1} + \dots + 1 \cdot {(\sqrt{a})}^{0} \cdot {(\sqrt{b})}^{12}$

Step 4

Simplify the polynomial result.

$a^{6} + 12 a^{5} \sqrt{b a} + 66 a^{5} b + 220 a^{4} b \sqrt{a b} + 495 a^{4} b^{2} + 792 a^{3} b^{2} \sqrt{a b} + 924 a^{3} b^{3} + 792 a^{2} b^{3} \sqrt{a b} + 495 a^{2} b^{4} + 220 a b^{4} \sqrt{a b} + 66 a b^{5} + 12 b^{5} \sqrt{a b} + b^{6}$

Expand Using the Binomial Theorem ( square root of a+ square root of b)^12

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