Question-183864

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

s i r, s h o w w o r k i n g s

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})

s i m i l a r l y

{(\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

t h a n k s s i r

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}]

δ = \sqrt{\frac{5}{11}}

k = 11^{4} [{(1 + δ)}^{8} + {(1 - δ)}^{8}]

= 11^{4} [2 (1 + (\begin{matrix} 8 \\ 2 \end{matrix}) δ^{2} + (\begin{matrix} 8 \\ 4 \end{matrix}) δ^{4} + (\begin{matrix} 8 \\ 6 \end{matrix}) δ^{6} + δ^{8})]

= 11^{4} [2 (1 + \frac{5}{11} (28 + 70 \frac{5}{11} + 28 \frac{5^{2}}{11^{2}} + \frac{5^{3}}{11^{3}})]

= 903 712

Commented by Michaelfaraday last updated on 31/Dec/22

t h a n k s s i r

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}, a b = 11 - 5 = 6

a^{2} + b^{2} = {(a + b)}^{2} - 2 a b

= {(2 \sqrt{11})}^{2} - 2 (6) = 44 - 12 = 10

a^{4} + b^{4} = {(a^{2} + b^{2})}^{2} - 2 a^{2} b^{2}

= 32^{2} - 2 {(6)}^{2} = 952

a^{8} + b^{8} = {(a^{4} + b^{4})}^{2} - 2 a^{4} b^{4}

= 952^{2} - 2 {(6)}^{4} = 903712

Commented by Acem last updated on 31/Dec/22

A l s o c o o l

Commented by manxsol last updated on 31/Dec/22

v e r y v e r y g o o d

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

T han X A c e m & m a n x s o l s i r s!

Commented by Michaelfaraday last updated on 31/Dec/22

t h a n k s s i r

Answered by manxsol last updated on 31/Dec/22

{(\sqrt{11} + \sqrt{5})}^{2} + {(\sqrt{11} - \sqrt{5})}^{2} = 2 (16) = 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 {(6)}^{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

t h a n k s s i r