Answer
Comment
Share
Q)

# Find $(x+1)^6 + (x – 1)^6$. Hence, evaluate $( \sqrt2 + 1)^6+(\sqrt 2 –1)^6$.

$\begin{array}{1 1}198 \\ 150 \\ 200 \\ 308 \end{array}$

## 1 Answer

Comment
A)
Need homework help? Click here.
Toolbox:
• Express the given number as the sum or difference of two numbers whose powers are easier to evaluate. Then use binomial theorem as follows:
• $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n, where b^0 = 1 = a^{n-n} We can now use Binomial Theorem to evaluate (x+1)^6 + (x – 1)^6. (x+1)^6 = \;^6\large C$$_0\;x^61^0 + \;^6\large C$$_1\;x^51^1+.... (x-1)^6 = \;^6\large C$$_0\;x^61^0 - \;^6\large C$$_1\;x^51^1+.... (x+1)^6 + (x – 1)^6 = \;^6\large C$$_0\;x^61^0 + \;^6\large C$$_1\;x^51^1+.... + ( \;^6\large C$$_0\;x^61^0 - \;^6\large C$$_1\;x^51^1+....) \qquad = 2 ( \;^6\large C$$_0\;x^6 + \;^6\large C$$_2\;x^4 + \;^6\large C$$_4\;x^2 + \;^6\large C$$_6)$
$\qquad = 2 (x^6+15x^4+15x^2+1)$
Therefore $( \sqrt2 + 1)^6+(\sqrt 2 –1)^6 = 2 (\sqrt{2}^6+15\;\sqrt{2}^4+15\;\sqrt{2}^2+1)$
$\qquad = 2(8+60+30+1)$
$\qquad = 2 (99)$
$\qquad = 198$