We obtain the required equation, by replacing the co-efficients in the reverse order, as \(5 x^{4}-4 x^{3}+7 x^{2}-5 x+1=0\)

To obtain the required equation, we have to multiply the co-efficients of  \(x^{3}, x^{2}, x\) and 1 by 1, 3, 3² and 3³ respectively. 
Thus \(x^{3}-3 x^{2}+9 x+27=0\) is the desired equation.

Let the roots of the given equation be a - d, a, a + d.
Then \(\mathrm{S}_{1}=a-d+a+a+d=3 a=-p \Rightarrow a=-\frac{p}{2}\)
Since a is a root, it satisfies the given polynomial
\(\Rightarrow\left(-\frac{p}{3}\right)^{3}+p\left(-\frac{p}{3}\right)^{2}+q\left(-\frac{p}{3}\right)+r=0\)
On simplification, we obtain \(2 p^{3}-9 p q+27 r=0\)

Since quadratic surds occur in pairs as roots, -\(\sqrt 3\) is also a root.
Since complex roots occur in conjugate pairs, 1+ 2i is also a root of the required polynomial equation. Therefore the desired equation is given by
\((x-\sqrt{3})(x+\sqrt{3})(x-(1-2 i))(x-(1+2 i))=0\)
\(\text { i.e., } x^{4}-2 x^{3}+2 x^{2}+6 x-15=0 \)

Let p(x) = x⁷-6x⁶+7x⁵+5x²+2x+2
It has only one change of sign. Now,
p(-x) = (-x)⁷ -6(-x)⁶ +7(-x)⁵ +5(-x)² +2(-x)+2
= -x⁷ -6x⁶ -7x⁵ + 5x² - 2x + 2
It has two, change of sign.
Hence, p(x) has one positive root and has at least two negative roots.

We have \(x=\sqrt { 2+x } \)
\(\Rightarrow { x }^{ 2 }=2+x \Rightarrow { x }^{ 2 }-x-2=0\)
\(\Rightarrow x=\frac { 1\pm \sqrt { 1+8 } }{ 2 } \Rightarrow x=\frac { 1\pm 3 }{ 2 } \)
\(\Rightarrow x=\frac { 1+3 }{ 2 } ,\frac { 1-3 }{ 2 } \Rightarrow x=2,-1\)
Also x>0, we get x = 2

The quadratic equation 3x² + 2(a² + 1)x + (a² - 3a + 2)
Will have two roots of opposite sign if it has real roots and the product of the roots is negative.
⇒ 4(a²+1)²-12(a²-3a+2)\(\ge\) 0 and \(\frac { { a }^{ 2 }-3a+2 }{ 3 } <0\)
Both of these conditions are true if 
⇒ a²-3a+2< 0
⇒ (a-1)(a-2)< 0
⇒ 1<a<2

Let ∝, β are the roots of the equation
Sum of the roots \(\alpha +\beta =\frac { -b }{ a } \)
 \(=\frac { [-(a-2)] }{ 1 } =a-2\)
Product of the roots \(=\alpha \beta =\frac { c }{ a } \)
\(=\frac { -(a+1) }{ 1 } =-(a+1)\)
we have \({ \alpha }^{ 2 }{ \beta }^{ 2 }=({ \alpha +\beta ) }^{ 2 }-2\alpha \beta \)
\(={ (a-2) }^{ 2 }+2(a+1)\)
\(={ a }^{ 2 }-4a+4+2a+2\)
\(=(a-1{ ) }^{ 2 }+5\)
Thus \(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }\) is least if a = 1

Sum of the roots = sin ∝ + cos ∝ = \(\frac{-b}{a}\)
Product of the roots = sin ∝ cos ∝ = \(\frac{c}{a}\)
Now 1 = cos²∝ + sin² ∝
= (sin ∝ +cos ∝)² - 2 sin ∝ cos ∝
\(1=\frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } \Rightarrow 1=\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \)
⇒ a² = b² - 2ac ⇒ a² + 2ac = b²
Adding c² both sides, a² +2ac+c² = b²+c²
⇒ (a+c)² = b² + c²

Here ∝ = 2, β = -2 and ૪ = 4
x³-(2-2+4)x²+(- 4 - 8 + 8x)x-(2)(-2)(4) = 0
⇒x³- 4x²- 4x+16 = 0

Let that number be x \( \\ \therefore \sqrt [ 3 ]{ x } +x=6\)
\(\Rightarrow \sqrt [ 3 ]{ x } =6-x\)
Taking power 3 both sides we get.
\({ \left( { x }^{ \frac { 1 }{ 3 } } \right) }^{ 3 }={ (6-x) }^{ 3 }\)
\(x={ 6 }^{ 3 }-3({ 6 }^{ 2 })x+3(6)({ x }^{ 2 })-{ x }^{ 3 }\)
\([\because { (a-b) }^{ 3 }={ a }^{ 3 }-3{ a }^{ 2 }b+{ 3ab }^{ 2 }-{ b }^{ 3 }]\)
\(\Rightarrow x=216-108x+18{ x }^{ 2 }-{ x }^{ 3 }\)
\(\Rightarrow { x }^{ 3 }+108x-18x-216+x=0\)
⇒ x³-18x²+109x -216 = 0. Which is the required mathematical problem.