Put 5+2√6 = y; Then 5-2√6 = \(\frac{1}{y}\)
∴ The given equataion becomes, 
\({ y }^{ { x }^{ 2 }-3 }+{ \left( \frac { 1 }{ y } \right) }^{ { x }^{ 2 }-3 }=10\)
Put  \({ y }^{ { x }^{ 2 }-3 }=z ...(1)\)
\(\Rightarrow z+\frac { 1 }{ z } =10\Rightarrow { z }^{ 2 }-10z+1=0\)
\(\Rightarrow z=\frac { 10\pm \sqrt { 100-4 } }{ 2 } =5\pm 2\sqrt { 6 } \)
\(\therefore (5+2\sqrt { 6 } )^{ { x }^{ 2 }-3 }=5\pm 2\sqrt { 6 } ={ (5\pm 2\sqrt { 6 } ) }^{ \pm 1 } \)
⇒x²- 3 = 1 or x²- 3 = -1
⇒ x² = 4 or x² = 2
⇒ x = 土2 or x = 土 √2
∴ The roots are 2,  -2, √2, -√2

Put y = \(\frac { x-1+x-5 }{ 2 } =-3\)
⇒ x = y + 3
∴ (x-1)⁴+(x-5)⁴ = 82
⇒ (y+3-1)⁴+(y+3-5)⁴ = 82
⇒ (y+2)⁴+(y-2)⁴ = 82
⇒ 2(y⁴+24y²+16) = 82
⇒ y⁴+24y²+16 = 41
⇒ y⁴+24y²-25 = 0
⇒ (y²+25)(y²-1) = 0
⇒ y = 土5i, y = 士1
∴ x = 3土5i, 4, 2.

The given equation can be written as
\({ 2 }^{ z }\left( 1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) ={ 7 }^{ x }\left( 1+\frac { 1 }{ 7 } +\frac { 1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 8+4+2 }{ 8 } \right) ={ 7 }^{ x }\left( \frac { 49+7+1 }{ 49 } \right) \)
\(\Rightarrow { 2 }^{ x }\left( \frac { 7 }{ 4 } \right) ={ 7 }^{ x }\left( \frac { 57 }{ 49 } \right) \Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } =\frac { { 7 }^{ x } }{ { 2 }^{ x } } \)
\(\Rightarrow \frac { 7 }{ 4 } \times \frac { 49 }{ 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\Rightarrow \frac { { 7 }^{ 3 } }{ 4\times 57 } ={ \left( \frac { 7 }{ 4 } \right) }^{ x }\)
\(\Rightarrow xlog\left( \frac { 7 }{ 4 } \right) =3log\ 7-log4-log57\)
\(\Rightarrow x=\frac { 3log7-log4-log57 }{ log\left( \frac { 7 }{ 2 } \right) } \)

Since x²- y² is even, either both x and y are even or both x and y are odd.
In any case both (x + y) and (x - y) are even integers
∴ x² - y²= (x + y) (x +y) must be divisible by 4 But 4 does not divide 353702
∴ x² - y² = 353702 has no positive integral solutions.

Given sin (e^x) -5^x + 5^-x
We have \({ 5 }^{ x }+{ 5 }^{ -x }={ ({ 5 }^{ \frac { x }{ 2 } }-{ 5 }^{ \frac { -x }{ 2 } }) }^{ 2 }+2\ge 2\)
If sin (e^x) -5^x + 5^-x has a solution
We get sin (e^x) ≥ 2 which i not possible for any
real x as |sin 0|≤ 1 for all θ∈ R.
∴ Sin (e^x) = 5^X + 5^-x has no solution.