2-2-8-A-B-Find-A-B-

Question Number 215439 by hardmath last updated on 06/Jan/25

{(2 + \sqrt{2})}^{8} = \sqrt{A} + \sqrt{B}

Find : A - B = ?

Answered by Rasheed.Sindhi last updated on 07/Jan/25

I f z i s a b i n o m i a l s u r d a n d n i s

n a t u r a l

\overset{―}{(z^{n})} = {(\overset{―}{z})}^{n}

I s t h i s c o r r e c t ?

{(2 + \sqrt{2})}^{8} = \sqrt{A} + \sqrt{B} \dots (i)

\Rightarrow {(2 - \sqrt{2})}^{8} = \sqrt{A} - \sqrt{B} \dots (i i)

(i) \times (i i) :

{(2 + \sqrt{2})}^{8} {(2 - \sqrt{2})}^{8} = A - B

{(2 + \sqrt{2}) (2 + \sqrt{2})}^{8} = A - B

{(4 - 2)}^{8} = A - B

A - B = 256

Commented by TonyCWX08 last updated on 07/Jan/25

Y e a h .

Commented by mr W last updated on 07/Jan/25

r i g h t!

g e n e r a l l y f o r a, b, c, n \in N :

{(a + b \sqrt{c})}^{n}

= \underset{k = 0}{\sum^{n}} C_{k}^{n} a^{n - k} {(b \sqrt{c})}^{k}

= \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋}} C_{2 r}^{n} a^{n - 2 r} {(b \sqrt{c})}^{2 r} + \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋ - 1}} C_{2 r + 1}^{n} a^{n - 2 r + 1} {(b \sqrt{c})}^{2 r + 1}

= \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋}} C_{2 r}^{n} a^{n - 2 r} b^{2 r} c^{r} + \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋ - 1}} C_{2 r + 1}^{n} a^{n - 2 r + 1} b^{2 r + 1} c^{r} \sqrt{c}

= \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋}} C_{2 r}^{n} a^{n - 2 r} b^{2 r} c^{r} + (\underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋ - 1}} C_{2 r + 1}^{n} a^{n - 2 r + 1} b^{2 r + 1} c^{r}) \sqrt{c}

= A + B \sqrt{c}

s i m i l a r l y

{(a - b \sqrt{c})}^{n}

= \underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋}} C_{2 r}^{n} a^{n - 2 r} b^{2 r} c^{r} - (\underset{r = 0}{\sum^{⌊ \frac{n}{2} ⌋ - 1}} C_{2 r + 1}^{n} a^{n - 2 r + 1} b^{2 r + 1} c^{r}) \sqrt{c}

= A - B \sqrt{c}

Commented by Rasheed.Sindhi last updated on 07/Jan/25

Than k s sirs!

Answered by TonyCWX08 last updated on 07/Jan/25

{(2 + \sqrt{2})}^{2} = 4 + 4 \sqrt{2} + 2 = 6 + 4 \sqrt{2}

{(6 + 4 \sqrt{2})}^{2} = 36 + 48 \sqrt{2} + 32 = 68 + 48 \sqrt{2}

{(68 + 48 \sqrt{2})}^{2} = 4624 + 6528 \sqrt{2} + 4608 = 9232 + 6528 \sqrt{2}

= \sqrt{9232^{2}} + \sqrt{2 {(6528)}^{2}}

A - B = 9232^{2} - 2 {(6528)}^{2} = 256

Commented by hardmath last updated on 07/Jan/25

{(2 + \sqrt{2})}^{8} \dots

Commented by TonyCWX08 last updated on 07/Jan/25

{(2 + \sqrt{2})}^{8} = {({({(2 + \sqrt{2})}^{2})}^{2})}^{2}