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.

Given ∝, β and ૪ are the roots of ax³ + bx² + cx + d = 0
\(\therefore \alpha +\beta +\gamma =\frac { -b }{ a } \)
\(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \)
\(\alpha \beta \gamma =\frac { -d }{ a } \)
Now, \(\sum { \frac { \alpha }{ \beta \gamma } } =\frac { \alpha }{ \beta \gamma } +\frac { \beta }{ \gamma \alpha } +\frac { \gamma }{ \alpha \beta } \)
\(=\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } }{ \alpha \beta \gamma } \)
\(=\frac { { (\alpha +\beta +\gamma ) }^{ 2 }-2(\alpha \beta +\beta \gamma +\gamma \alpha ) }{ \alpha \beta \gamma } \)
\(=\frac { { \left( -\frac { b }{ a } \right) }^{ 2 }-2\left( \frac { c }{ a } \right) }{ -\frac { d }{ a } } \)
\(=\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } }{ -\frac { d }{ a } } \Rightarrow \frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \times \frac { -a }{ d } \)
\(\therefore \frac { \sum { \alpha } }{ \beta \gamma } =-\frac { \left( { b }^{ 2 }-2ac \right) }{- ad } =\frac { 2ac-{ b }^{ 2 } }{ ad } \)

Let the roots form a GP be \(\frac{a}{\lambda}\), a, a\(\lambda\)
Product of the roots \(\frac{a}{\lambda} \times a \times a \lambda=\frac{24}{3}=8\)
a³ = 8
a = 2
Sum of the roots, \(\frac{a}{\lambda}+a+a \lambda=\frac{26}{3}\)
\( a\left(\frac{1}{\lambda}+1+\lambda\right) =\frac{26}{3} \)
\(2\left(\frac{1+\lambda+\lambda^{2}}{\lambda}\right) =\frac{26}{3} \)
\( 3+3 \lambda+3 \lambda^{2} =13 \lambda \)
\(3 \lambda^{2}+3 \lambda-13 \lambda+3 =0 \)
\(3 \lambda^{2}-10 \lambda+3 =0\)
\( (\lambda-3)(3 \lambda-1)=0 \)
\( \lambda=3 \text { (or) } \lambda=\frac{1}{3}\)
\( \lambda=3, \quad \frac{a}{\lambda}=\frac{2}{3} , \)
\( a \lambda=2(3)=6 \)
If \( \lambda=\frac{1}{3}, \frac{a}{\lambda}=\frac{2}{\frac{1}{3}}=6 \)
If \( a \lambda=2\left(\frac{1}{3}\right)=\frac{2}{3} \)
Roots are \(\frac{a}{\lambda},\ a,\ a \lambda\)
If \(\lambda\) = 3 roots are \(\frac{2}{3}\), 2, 6
If \(\lambda=\frac{1}{3}\) roots are 6, 2, \(\frac{2}{3}\)

Here, a = 9, b = - 36, c = 44, d = -16
Since the roots form an arithmetic progression,
Let the roots be a - d, a and a + d
Sum of the roots \(=\frac { -b }{ a } \)
\(\Rightarrow (a-d)+(a)+(a+d)=\frac { -(-36) }{ 9 } =4\)
\(\Rightarrow 3a=4\Rightarrow a=\frac { 4 }{ 3 } \)
and product of the roots \(=\frac { -d }{ a } \)
\(=\frac { -(-16) }{ 9 } =\frac { 16 }{ 9 } \)
\(\Rightarrow (a-d)(a)(a+d)=\frac { 16 }{ 9 } \)
\(\left( { a }^{ 2 }-{ d }^{ 2 } \right) (a)=\frac { 16 }{ 9 } \)
\(\left( \frac { 16 }{ 9 } -{ d }^{ 2 } \right) \left( \frac { 4 }{ 3 } \right) =\frac { 16 }{ 9 } \ [\because a=\frac { 4 }{ 3 } ]\)
\(\frac { 16 }{ 9 } -{ d }^{ 2 }=\frac { 16 }{ 9 } \times \frac { 3 }{ 4 } =\frac { 4 }{ 3 } \)
\(\frac { 16 }{ 9 } -\frac { 4 }{ 3 } ={ d }^{ 2 }=\frac { 16 }{ 9 } \times \frac { 3 }{ 4 } =\frac { 4 }{ 3 } \)
\(\frac { 16 }{ 9 } -\frac { 4 }{ 3 } ={ d }^{ 2 }\)
\(\Rightarrow { d }^{ 2 }=\frac { 16-12 }{ 9 } =\frac { 4 }{ 9 } \)
\(\Rightarrow d=\pm \sqrt { \frac { 4 }{ 9 } } =\frac { 2 }{ 3 } \)
∴ The roots are a-d, a, a+d
\(\Rightarrow \frac { 4 }{ 3 } -\frac { 2 }{ 3 } ,\frac { 4 }{ 3 } ,\frac { 4 }{ 3 } +\frac { 2 }{ 3 } \Rightarrow \frac { 2 }{ 3 } ,\frac { 4 }{ 3 } ,2\)

Given equation are \({ x }^{ 2 }px+q=0\)  ............(1)
and \({ x }^{ 2 }+p'x+q'=0\)    ..........(2)
Let ∝ be the common root for (1) and (2)
∴ ∝² + p∝ + q = 0   .........(3)
and ∝²+ p'∝ + q' = 0    .............(4)
Solving (3) and (4) by cross multiplication method we get
p  q 1 p
p' q' 1 p'
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \)
consider \(\frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } \)
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ \alpha } =\frac { pq'-p'q }{ q-q' } \)
\(\alpha =\frac { pq'-p'q }{ q-q } \)
Consider \(\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \Rightarrow \alpha =\frac { q-q' }{ p'-p } \)
Hence its roots are \(\frac { pq'-p'q }{ q-q' } or\quad \frac { q-q' }{ p'-p } \)

Given p, q are the roots of lx² + nx + n = 0 
\(p+q=\frac { -b }{ a } =\frac { -n }{ l } ...(1)\)
\(pq=\frac { c }{ a } =\frac { n }{ l } ...(2)\)
L.H.S 
\( \sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}} =\frac{\sqrt{p}}{\sqrt{q}}+\frac{\sqrt{q}}{\sqrt{p}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{p+q}{\sqrt{p} \sqrt{q}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\( =\frac{-\frac{n}{l}}{\sqrt{p q}}+\frac{\sqrt{n}}{\sqrt{l}}=\frac{-\frac{n}{l}}{\frac{\sqrt{n}}{\sqrt{l}}}+\frac{\sqrt{n}}{\sqrt{l}} \)
\(=\frac{-\frac{n}{l}+\frac{n}{l}}{\sqrt{\frac{n}{l}}}=0\) R.H.S