\( \alpha \beta=\frac{\alpha \beta \gamma}{\gamma}=\frac{-c}{\gamma} \)
put \( y=\frac{-c}{x} \Rightarrow x=\frac{-c}{y} \)
Hence the given equation becomes
\(\left(\frac{-c}{y}\right)^{3}+a\left(\frac{-c}{y}\right)^{2}+b\left(\frac{-c}{y}\right)+c=0\)
\(\text { i.e., } y^{3}-b y^{2}+a c y-c^{2}=0 \text {, }\)which is the required equation.

Let the given equation be
\(x^{4}+P_{1} x^{3}+P_{2} x^{2}+P_{3} x+P_{4}=0\)
Here P₁ = P₂ = 0, P₃ = 2 and P₄ = 3
By Newton's theorem, \(S_{3}+S_{2} P_{1}+S_{1} P_{2}+3 P_{3}=0\)
\(\text { i.e., } \mathrm{S}_{3}+0+0+3.2=0 \)
\(\Rightarrow \mathrm{S}_{3}=-6 \)
i.e., sum of the cubes of the roots of \(x^{4}+2 x+3=0\) is -6

Let the roots are in H.P. Then their reciprocals are in A.P.
\( \frac{15}{x^{3}}-\frac{23}{x^{2}}+\frac{9}{x}-1=0 \)
\( 15-23 x+9 x^{2}-x^{3}=0 \)
\(x^{3}-9 x^{2}+23 x-15=0\)
Since the roots are in A.P. Assume that the roots are in the form a-d, a, a + d
\( \Sigma_{1}=a-d+a+a+d =9 \)
\(3 a =9 \)
\(a =3 \)
\(\Sigma_{3}=(a-d)(a)(a+d) =15 \)
\(a\left(a^{2}-d^{2}\right) =15 \)
\(3\left(9-d^{2}\right) =15 \)
\(9-d^{2} =\frac{15}{3} \)
\(9-d^{2} =5 \\d^{2} =4 \)
\(d =\pm 2 \)
a = 3, d = 2 then roots are 1, 3, 5
a = 3, d = -2 then roots are 5, 3, 1

Let the roots be \(\frac{a}{r}, a, a r\)
Then \(\frac{a}{r} \times a \times a r=a^{3}=\frac{8}{27} \Rightarrow a=\frac{2}{3}\)
Since \(a=\frac{2}{3}\) is a root, \(\left(x-\frac{2}{3}\right)\) is a factor
On division, the other factor of the polynomial is \(27 x^{2}+60 x+12\)
Its roots are \(\frac{-60 \pm \sqrt{60^{2}-4 \times 27 \times 12}}{2 \times 27}=-\frac{2}{9} \text { or }-2\)
Hence the roots of the given polynomial equation are \(-\frac{2}{9},-2, \frac{2}{3}\)

Since quadratic surds occur in conjugate pairs as roots of a polynomial equation, \(1+\sqrt{5} \) is also a root of the given polynomial.
\( \Rightarrow \quad[x-(1-\sqrt{5})][x-(1+\sqrt{5})]=(x-1)^{2}-5 \)
\( =x^{2}-2 x-4 \) is a factor
Dividing the given polynomial by this factor, we obtain the other factor as \( x^{2}-3 x+2\).
Also, \(x^{2}-3 x+2=(x-2)(x-1)\)
Thus the roots of the given polynomial equation are \(1+\sqrt{5}, 1-\sqrt{5}, 1,2\)

Given that \(2+i \sqrt{3}\) is a root of \(x^{4}-4 x^{2}+8 x+35=0\).
Since complex roots occurs in conjugate pairs, \(2-i \sqrt{3}\) is also a root of it.
\(\Rightarrow[x-(2+i \sqrt{3})][x-(2-i \sqrt{3})]=(x-2)^{2}+3\)
\(=x^{2}-4 x+7\)  is a factor of the given polynomial.
Dividing the given polynomial by this factor, we obtain the other factor as \(x^{2}+4 x+5\)
The roots of x²+ 4x + 5 = 0 are given by \(\frac{-4 \pm \sqrt{16-20}}{2}=-2 \pm i\)
Hence the roots of the given polynomial are
\(2+i \sqrt{3}, 2-i \sqrt{3},-2+i \text { and }-2-i\)

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.