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Question Number 183864 by Michaelfaraday last updated on 31/Dec/22

Commented by MJS_new last updated on 31/Dec/22

$$b)$$

Commented by Michaelfaraday last updated on 31/Dec/22

$$sir,showworkings$$

Answered by mr W last updated on 31/Dec/22

$${(\sqrt{11}+\sqrt{5})}^2=16+2\sqrt{55}$$$${(\sqrt{11}+\sqrt{5})}^4=4(119+16\sqrt{55})$$$${(\sqrt{11}+\sqrt{5})}^8=16({119}^2+{16}^2\times 55+2\times 119\times 16\sqrt{55})$$$$similarly$$$${(\sqrt{11}-\sqrt{5})}^8=16({119}^2+{16}^2\times 55-2\times 119\times 16\sqrt{55})$$$${(\sqrt{11}+\sqrt{5})}^8+{(\sqrt{11}-\sqrt{5})}^8=2\times 16({119}^2+{16}^2\times 55)$$$$=903712✓$$

Commented by Michaelfaraday last updated on 31/Dec/22

$$thankssir$$

Answered by Acem last updated on 31/Dec/22

$$k={(\sqrt{11}+\sqrt{5})}^8+{(\sqrt{11}-\sqrt{5})}^8={11}^4[{(1+\sqrt{\frac{5}{11}})}^8+{(1-\sqrt{\frac{5}{11}})}^8]$$$$\delta =\sqrt{\frac{5}{11}}$$$$k={11}^4[{(1+\delta )}^8+{(1-\delta )}^8]$$$$={11}^4\left[2(1+\left(\begin{array}{c}8\\ 2\end{array}\right){\delta }^2+\left(\begin{array}{c}8\\ 4\end{array}\right){\delta }^4+\left(\begin{array}{c}8\\ 6\end{array}\right){\delta }^6+{\delta }^8)\right]$$$$={11}^4[2(1+\frac{5}{11}(28+70\frac{5}{11}+28\frac{5^2}{{11}^2}+\frac{5^3}{{11}^3})]$$$$=903712$$$$$$

Commented by Michaelfaraday last updated on 31/Dec/22

$$thankssir$$

Answered by Rasheed.Sindhi last updated on 31/Dec/22

$$a=\sqrt{11}+\sqrt{5},b=\sqrt{11}-\sqrt{5}$$$$a+b=2\sqrt{11},ab=11-5=6$$$$a^2+b^2={(a+b)}^2-2ab$$$$={\left(2\sqrt{11}\right)}^2-2\left(6\right)=44-12=10$$$$a^4+b^4={(a^2+b^2)}^2-2a^2{b}^2$$$$={32}^2-2{\left(6\right)}^2=952$$$$a^8+b^8={(a^4+b^4)}^2-2a^4{b}^4$$$$={952}^2-2{\left(6\right)}^4=903712$$

Commented by Acem last updated on 31/Dec/22

$$Alsocool$$

Commented by manxsol last updated on 31/Dec/22

$$veryverygood$$

Commented by Rasheed.Sindhi last updated on 31/Dec/22

$$\mathbb{T}\mathbf{han}\mathbb{X}Acem\mathrm{\&}manxsolsirs!$$

Commented by Michaelfaraday last updated on 31/Dec/22

$$thankssir$$

Answered by manxsol last updated on 31/Dec/22

$${(\sqrt{11}+\sqrt{5})}^2+{(\sqrt{11}-\sqrt{5})}^2=2\left(16\right)=32$$$${(a+b)}^2+{(a-b)}^2=2(a^2+b^2)$$$${(\sqrt{11}+\sqrt{5})}^4+{(\sqrt{11}-\sqrt{5})}^4+2{(11-5)}^2={32}^2$$$$2{(\sqrt{11}+\sqrt{5})}^2{(\sqrt{11}+\sqrt{5})}^2=2{[\sqrt{11}{}^2+\sqrt{5}{}^2]}^2=2{(11-5)}^2$$$${(\sqrt{11}+\sqrt{5})}^4+{(\sqrt{11}-\sqrt{5})}^4={32}^2-2\times 6^2=952$$$${(\sqrt{11}+\sqrt{5})}^8+{(\sqrt{11}-\sqrt{5})}^8+2{(11-5)}^4={952}^2$$$${(\sqrt{11}+\sqrt{5})}^8+{(\sqrt{11}-\sqrt{5})}^8={952}^2-2{\left(6\right)}^4$$$${(\sqrt{11}+\sqrt{5})}^8+{(\sqrt{11}-\sqrt{5})}^8=$$$${952}^2-2\times 6^4$$$$903712.0$$$$$$

Commented by Michaelfaraday last updated on 31/Dec/22

$$thankssir$$

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