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

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Question Number 215439 by hardmath last updated on 06/Jan/25
( 2 + (√2) )^8 = (√A) + (√B) Find: A−B = ?
$$\left(\:\mathrm{2}\:\:+\:\:\sqrt{\mathrm{2}}\:\right)^{\mathrm{8}} \:=\:\:\sqrt{\mathrm{A}}\:\:+\:\:\sqrt{\mathrm{B}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{A}−\mathrm{B}\:=\:? \\ $$
Answered by Rasheed.Sindhi last updated on 07/Jan/25
If z is a binomial surd and n is natural (z^n )^(−) =(z^(−) )^n Is this correct? ( 2 + (√2) )^8 = (√A) + (√B) ...(i) ⇒( 2 − (√2) )^8 = (√A) − (√B) ...(ii) (i)×(ii): ( 2 + (√2) )^8 ( 2 − (√2) )^8 =A−B {(2+(√2) )(2+(√2) )}^8 =A−B (4−2)^8 =A−B A−B=256
$${If}\:{z}\:{is}\:{a}\:{binomial}\:{surd}\:{and}\:{n}\:{is} \\ $$$${natural}\: \\ $$$$\:\:\:\overline {\left({z}^{{n}} \right)}=\left(\overline {{z}}\right)^{{n}} \\ $$$${Is}\:{this}\:{correct}? \\ $$$$\: \\ $$$$\:\:\:\:\:\left(\:\mathrm{2}\:\:+\:\:\sqrt{\mathrm{2}}\:\right)^{\mathrm{8}} \:=\:\:\sqrt{\mathrm{A}}\:\:+\:\:\sqrt{\mathrm{B}}\:…\left({i}\right) \\ $$$$\Rightarrow\left(\:\mathrm{2}\:\:−\:\:\sqrt{\mathrm{2}}\:\right)^{\mathrm{8}} \:=\:\:\sqrt{\mathrm{A}}\:\:−\:\:\sqrt{\mathrm{B}}\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\:\:\:\:\left(\:\mathrm{2}\:\:+\:\:\sqrt{\mathrm{2}}\:\right)^{\mathrm{8}} \left(\:\mathrm{2}\:\:−\:\:\sqrt{\mathrm{2}}\:\right)^{\mathrm{8}} =\mathrm{A}−\mathrm{B} \\ $$$$\:\:\:\:\left\{\left(\mathrm{2}+\sqrt{\mathrm{2}}\:\right)\left(\mathrm{2}+\sqrt{\mathrm{2}}\:\right)\right\}^{\mathrm{8}} =\mathrm{A}−\mathrm{B} \\ $$$$\:\:\:\:\:\left(\mathrm{4}−\mathrm{2}\right)^{\mathrm{8}} =\mathrm{A}−\mathrm{B} \\ $$$$\:\:\:\:\:\:\:\mathrm{A}−\mathrm{B}=\mathrm{256} \\ $$
Commented by TonyCWX08 last updated on 07/Jan/25
Yeah.
$${Yeah}. \\ $$
Commented by mr W last updated on 07/Jan/25
right! generally for a,b,c,n∈N: (a+b(√c))^n =Σ_(k=0) ^n C_k ^n a^(n−k) (b(√c))^k =Σ_(r=0) ^(⌊(n/2)⌋) C_(2r) ^n a^(n−2r) (b(√c))^(2r) +Σ_(r=0) ^(⌊(n/2)⌋−1) C_(2r+1) ^n a^(n−2r+1) (b(√c))^(2r+1) =Σ_(r=0) ^(⌊(n/2)⌋) C_(2r) ^n a^(n−2r) b^(2r) c^r +Σ_(r=0) ^(⌊(n/2)⌋−1) C_(2r+1) ^n a^(n−2r+1) b^(2r+1) c^r (√c) =Σ_(r=0) ^(⌊(n/2)⌋) C_(2r) ^n a^(n−2r) b^(2r) c^r +(Σ_(r=0) ^(⌊(n/2)⌋−1) C_(2r+1) ^n a^(n−2r+1) b^(2r+1) c^r )(√c) =A+B(√c) similarly (a−b(√c))^n =Σ_(r=0) ^(⌊(n/2)⌋) C_(2r) ^n a^(n−2r) b^(2r) c^r −(Σ_(r=0) ^(⌊(n/2)⌋−1) C_(2r+1) ^n a^(n−2r+1) b^(2r+1) c^r )(√c) =A−B(√c)
$${right}!\: \\ $$$${generally}\:{for}\:{a},{b},{c},{n}\in{N}: \\ $$$$\left({a}+{b}\sqrt{{c}}\right)^{{n}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {a}^{{n}−{k}} \left({b}\sqrt{{c}}\right)^{{k}} \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\lfloor\frac{{n}}{\mathrm{2}}\rfloor} {\sum}}{C}_{\mathrm{2}{r}} ^{{n}} {a}^{{n}−\mathrm{2}{r}} \left({b}\sqrt{{c}}\right)^{\mathrm{2}{r}} +\underset{{r}=\mathrm{0}} {\overset{\lfloor\frac{{n}}{\mathrm{2}}\rfloor−\mathrm{1}} {\sum}}{C}_{\mathrm{2}{r}+\mathrm{1}} ^{{n}} {a}^{{n}−\mathrm{2}{r}+\mathrm{1}} \left({b}\sqrt{{c}}\right)^{\mathrm{2}{r}+\mathrm{1}} \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\lfloor\frac{{n}}{\mathrm{2}}\rfloor} {\sum}}{C}_{\mathrm{2}{r}} ^{{n}} {a}^{{n}−\mathrm{2}{r}} {b}^{\mathrm{2}{r}} {c}^{{r}} +\underset{{r}=\mathrm{0}} {\overset{\lfloor\frac{{n}}{\mathrm{2}}\rfloor−\mathrm{1}} {\sum}}{C}_{\mathrm{2}{r}+\mathrm{1}} ^{{n}} {a}^{{n}−\mathrm{2}{r}+\mathrm{1}} {b}^{\mathrm{2}{r}+\mathrm{1}} {c}^{{r}} \sqrt{{c}} \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\lfloor\frac{{n}}{\mathrm{2}}\rfloor} {\sum}}{C}_{\mathrm{2}{r}} ^{{n}} {a}^{{n}−\mathrm{2}{r}} {b}^{\mathrm{2}{r}} {c}^{{r}} +\left(\underset{{r}=\mathrm{0}} {\overset{\lfloor\frac{{n}}{\mathrm{2}}\rfloor−\mathrm{1}} {\sum}}{C}_{\mathrm{2}{r}+\mathrm{1}} ^{{n}} {a}^{{n}−\mathrm{2}{r}+\mathrm{1}} {b}^{\mathrm{2}{r}+\mathrm{1}} {c}^{{r}} \right)\sqrt{{c}} \\ $$$$={A}+{B}\sqrt{{c}} \\ $$$${similarly} \\ $$$$\left({a}−{b}\sqrt{{c}}\right)^{{n}} \\ $$$$=\underset{{r}=\mathrm{0}} {\overset{\lfloor\frac{{n}}{\mathrm{2}}\rfloor} {\sum}}{C}_{\mathrm{2}{r}} ^{{n}} {a}^{{n}−\mathrm{2}{r}} {b}^{\mathrm{2}{r}} {c}^{{r}} −\left(\underset{{r}=\mathrm{0}} {\overset{\lfloor\frac{{n}}{\mathrm{2}}\rfloor−\mathrm{1}} {\sum}}{C}_{\mathrm{2}{r}+\mathrm{1}} ^{{n}} {a}^{{n}−\mathrm{2}{r}+\mathrm{1}} {b}^{\mathrm{2}{r}+\mathrm{1}} {c}^{{r}} \right)\sqrt{{c}} \\ $$$$={A}−{B}\sqrt{{c}} \\ $$
Commented by Rasheed.Sindhi last updated on 07/Jan/25
Thanks sirs!
$$\boldsymbol{\mathrm{Than}}\Bbbk\boldsymbol{\mathrm{s}}\:\:\boldsymbol{\mathrm{sirs}}! \\ $$
Answered by TonyCWX08 last updated on 07/Jan/25
(2+(√2))^2 =4+4(√2)+2=6+4(√2) (6+4(√2))^2 =36+48(√2)+32=68+48(√2) (68+48(√2))^2 =4624+6528(√2)+4608=9232+6528(√2) =(√(9232^2 ))+(√(2(6528)^2 )) A−B=9232^2 −2(6528)^2 =256
$$\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{4}+\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{2}=\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{36}+\mathrm{48}\sqrt{\mathrm{2}}+\mathrm{32}=\mathrm{68}+\mathrm{48}\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{68}+\mathrm{48}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{4624}+\mathrm{6528}\sqrt{\mathrm{2}}+\mathrm{4608}=\mathrm{9232}+\mathrm{6528}\sqrt{\mathrm{2}} \\ $$$$=\sqrt{\mathrm{9232}^{\mathrm{2}} }+\sqrt{\mathrm{2}\left(\mathrm{6528}\right)^{\mathrm{2}} } \\ $$$${A}−{B}=\mathrm{9232}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{6528}\right)^{\mathrm{2}} =\mathrm{256} \\ $$
Commented by hardmath last updated on 07/Jan/25
(2 + (√2))^8 ...
$$\left(\mathrm{2}\:+\:\sqrt{\mathrm{2}}\right)^{\mathrm{8}} \:… \\ $$
Commented by TonyCWX08 last updated on 07/Jan/25
(2+(√2))^8 =(((2+(√2))^2 )^2 )^2
$$\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{8}} =\left(\left(\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \right)^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$

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