Question

Simplify: $(\sqrt{2}+1)^{6}-(\sqrt{2}-1)^{6}$

• A. $140\sqrt{2}$
• B. 101
• C. $70\sqrt{2}$
• D. $120\sqrt{2}$

Answer 1

Answer: (A)

$140\sqrt{2}$

Answer 2

To simplify the expression $(\sqrt{2}+1)^{6}-(\sqrt{2}-1)^{6}$, we can use the binomial theorem.

Let $a = \sqrt{2}$ and $b = 1$. The expression is of the form $(a+b)^6 - (a-b)^6$.

The binomial expansion for $(a+b)^n$ is: $(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}a^0 b^n$

The binomial expansion for $(a-b)^n$ is: $(a-b)^n = \binom{n}{0}a^n b^0 - \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 - \dots + (-1)^n \binom{n}{n}a^0 b^n$

When we subtract $(a-b)^n$ from $(a+b)^n$, the terms with even powers of $b$ cancel out, and the terms with odd powers of $b$ are doubled: $(a+b)^n - (a-b)^n = 2 \left[ \binom{n}{1}a^{n-1}b^1 + \binom{n}{3}a^{n-3}b^3 + \binom{n}{5}a^{n-5}b^5 + \dots \right]$

In this case, $n=6$, $a=\sqrt{2}$, and $b=1$. So, the expression becomes: $(\sqrt{2}+1)^6 - (\sqrt{2}-1)^6 = 2 \left[ \binom{6}{1}(\sqrt{2})^{6-1}(1)^1 + \binom{6}{3}(\sqrt{2})^{6-3}(1)^3 + \binom{6}{5}(\sqrt{2})^{6-5}(1)^5 \right]$ $= 2 \left[ \binom{6}{1}(\sqrt{2})^5 + \binom{6}{3}(\sqrt{2})^3 + \binom{6}{5}(\sqrt{2})^1 \right]$

Now, we calculate the binomial coefficients and powers of $\sqrt{2}$: $\binom{6}{1} = 6$ $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ $\binom{6}{5} = \binom{6}{6-5} = \binom{6}{1} = 6$

$(\sqrt{2})^1 = \sqrt{2}$ $(\sqrt{2})^3 = (\sqrt{2})^2 \times \sqrt{2} = 2\sqrt{2}$ $(\sqrt{2})^5 = (\sqrt{2})^4 \times \sqrt{2} = ((\sqrt{2})^2)^2 \times \sqrt{2} = 2^2 \times \sqrt{2} = 4\sqrt{2}$

Substitute these values back into the expression: $= 2 \left[ 6(4\sqrt{2}) + 20(2\sqrt{2}) + 6(\sqrt{2}) \right]$ $= 2 \left[ 24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2} \right]$ $= 2 \left[ (24+40+6)\sqrt{2} \right]$ $= 2 \left[ 70\sqrt{2} \right]$ $= 140\sqrt{2}$