Expand Using the Binomial Theorem ( square root of x- square root of 2)^6

${(\sqrt{x} - \sqrt{2})}^{6}$

Step 1

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

$\sum_{k = 0}^{6} \frac{6!}{(6 - k)! k!} \cdot {(\sqrt{x})}^{6 - k} \cdot {(- \sqrt{2})}^{k}$

Step 2

Expand the summation.

$\frac{6!}{(6 - 0)! 0!} \cdot {(\sqrt{x})}^{6 - 0} \cdot {(- \sqrt{2})}^{0} + \frac{6!}{(6 - 1)! 1!} \cdot {(\sqrt{x})}^{6 - 1} \cdot {(- \sqrt{2})}^{1} + \dots + \frac{6!}{(6 - 6)! 6!} \cdot {(\sqrt{x})}^{6 - 6} \cdot {(- \sqrt{2})}^{6}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(\sqrt{x})}^{6} \cdot {(- \sqrt{2})}^{0} + 6 \cdot {(\sqrt{x})}^{5} \cdot {(- \sqrt{2})}^{1} + \dots + 1 \cdot {(\sqrt{x})}^{0} \cdot {(- \sqrt{2})}^{6}$

Step 4

Simplify the polynomial result.

$x^{3} - 6 x^{2} \sqrt{2 x} + 30 x^{2} - 40 x \sqrt{2 x} + 60 x - 24 \sqrt{2 x} + 8$

Expand Using the Binomial Theorem ( square root of x- square root of 2)^6

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