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Question Number 51235 by peter frank last updated on 25/Dec/18

$$Withoutusingtables,$$$$findthavalueof$$$$\frac{{(\sqrt{5}+2)}^6-{(\sqrt{5}-2)}^6}{8\sqrt{5}}$$

Answered by mr W last updated on 25/Dec/18

$$leta=\sqrt{5}+2$$$$\sqrt{5}-2=\frac{1}{\sqrt{5}+2}=\frac{1}{a}$$$$a^6-\frac{1}{a^6}=(a^3+\frac{1}{a^3})(a^3-\frac{1}{a^3})=(a+\frac{1}{a})(a^2+\frac{1}{a^2}-1)(a-\frac{1}{a})(a^2+\frac{1}{a^2}+1)$$$$=(a^2-\frac{1}{a^2})[{(a^2+\frac{1}{a^2})}^2-1]$$$$=\left(8\sqrt{5}\right)({18}^2-1)$$$$\Rightarrow \frac{{(\sqrt{5}+2)}^6-{(\sqrt{5}-2)}^6}{8\sqrt{5}}={18}^2-1$$$$=17\times 19=323$$

Commented by peter frank last updated on 25/Dec/18

$$muchrespectsirnicework$$

Commented by malwaan last updated on 25/Dec/18

$$\mathrm{great}\mathrm{work}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18

$${(a+b)}^6=a^6+6c_1{a}^{5}b+6c_2{a}^4{b}^2+6c_3{a}^3{b}^3+6c_4{a}^2{b}^4+6c_5{ab}^5+b^6$$$${(a-b)}^6=a^6-6c_1{a}^{5}b+6c_2{a}^4{b}^2-6c_3{a}^3{b}^3+6c_4{a}^2{b}^4-6c_5{ab}^5+b^6$$$${(a+b)}^6-{(a-b)}^6$$$$=2[6c_1{a}^{5}b+6c_3{a}^3{b}^3+6c_5{ab}^5]$$$$=2[6a^{5}b+20a^3{b}^3+6{ab}^5]$$$$=12{\left(\sqrt{5}\right)}^5\left(2\right)+40{\left(\sqrt{5}\right)}^3{\left(2\right)}^3+12\left(\sqrt{5}\right){\left(2\right)}^5$$$$=12\times 2\times 25\times \sqrt{5}+40\times 8\times 5\sqrt{5}+32\times 12\times \sqrt{5}$$$$=8\sqrt{5}(75+200+48)$$$$ansis75+200+48=323$$$$$$$$$$$$$$$$$$

Commented by peter frank last updated on 25/Dec/18

$$perfectsir.binomialexpansion$$$$ihesitatetoapplythat$$$$method$$

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