Since α, β, and γ are the roots of the equation x³+ px²+ qx + r = 0, we have
Σ₁ α + β + γ = -p and Σ₃ αβγ = -r
\(\Sigma \frac { 1 }{ \beta \gamma } =\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } +\frac { 1 }{ \alpha \beta } =\frac { \alpha +\beta +\gamma }{ \alpha \beta \gamma } =\frac { -p }{ -r } =\frac { p }{ r } \).

Given that the height of the tree is 12m.
Let x m be the standing part and (12 -x)m be the broken part.
Given \(x=\sqrt [ 3 ]{ 12-x } \)
\(\Rightarrow x=(12-x)^{ \frac { 1 }{ 3 } }\)


Taking power 3 both sides, we get 
⇒ x³ = 12-x
⇒ x³+ x -12 = 0
which is the required mathematical problem.

Let α, β, γ and δ be the roots of ax⁴+ bx³+ cx²+ dx + e = 0
Σ₁ = α + β + γ + δ = -\(\frac { b }{ a } \),
Σ₂ = αβ + αγ + αδ + βγ + βδ + γδ = \(\frac { c }{ a } \),
Σ₃ = αβγ + αβδ +  αγδ +  βγδ = -\(\frac { d }{ a } \)
Σ₄ = αβγδ =\(\frac { e }{ a } \)
We have to find α² + β² + γ² + δ² 
Applying the algebraic identity
(a+b+c+d)² ≡ a²+b²+c²+d²+2(ab+ac+ad+bc+bd+cd),
we get
α² + β² + γ² + δ² = (α + β + γ + δ)²-2(αβ + αγ + αδ + βγ + βδ + γδ)
= \(\left( \frac { b }{ a } \right) ^{ 2 }-2\left( \frac { c }{ a } \right) \)
= \(\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \).

6x⁴- 35x³+ 62x²- 35x + 6 = 0
This equation is type I even degree reciprocal equation.
Hence, it can be rewritten as
\(6\left( { x }^{ 2 }+\frac { 1 }{ x } \right) -35(x+\frac { 1 }{ x } )+62=0 ...(1)\)
putting \(x+\frac { 1 }{ x } =y\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2={ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ y }^{ 2 }-2\)
∴ (1) becomes as,
\(\Rightarrow 6({ y }^{ 2 }-2)-35y+62=0\)
\(\Rightarrow { 6y }^{ 2 }-12-35y+62=0\)
\(\Rightarrow { 6y }^{ 2 }-35y+50=0\)
\(\Rightarrow (3y-10)(2y-5)=0\)
\(\Rightarrow y=\frac { 10 }{ 3 } ,\frac { 5 }{ 2 } \)
Case (i) when \(y=\frac { 10 }{ 3 } ,x+\frac { 1 }{ x } =\frac { 10 }{ 3 } \)


\(\Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { 10 }{ 3 } \)
\(\Rightarrow { 3x }^{ 2 }-10x+3=10x\)
\(\Rightarrow { 3x }^{ 2 }-10x+3=0\)
\(\Rightarrow (x-3)(3x-1)=0\)
\(\Rightarrow x=3,\frac { 1 }{ 3 } \)
Case (ii) when \(y=\frac { 5 }{ 2 } ,x+\frac { 1 }{ x } =\frac { 5 }{ 2 } \Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { 5 }{ 2 } \)
\(\Rightarrow { 2x }^{ 2 }+2=5x\Rightarrow { 2x }^{ 2 }-5x+2=0\)
\(\Rightarrow (x-2)(2x-1)=0\)
Hence the roots are \(2,\frac { 1 }{ 2 } ,3,\frac { 1 }{ 3 } \)

Since the sum of the co-efficients is
2 - 9 + 10 - 3 = 12 - 12 = 0
⇒ x = 1 is a root of (x)
∴ (x - 1) is a factor of (x)
To find the other factor, let us divide f(x) by x-1 

[Using synthetic division]
f(x) = (x-1)(x - 3)(2x - 1) = 0
⇒x - 1 = 0, x - 3 = 0 or 2x - 1 = 0
⇒ x = 1, x = 3x, x = \(\frac{1}{2}\)
Hence the roots are 1, 3, \(\frac{1}{2}\).

The given equation is same as
(2x-3)(3x-2)(6x-1)(x-2)-5 = 0
After a computation, the above equation becomes
(6x²-13x+6)(6x²-13x+12)-5 = 0
By taking y = 6x²-13x, the above equation becomes
(y+6)(y+12)-5 = 0
which is same as
y²+18y+7 = 0
Solving this equation, we get y = −1 and y = −7.
Substituting the values of y in y = −6x²-13x, we get
6x²-13x+1 = 0
6x²-13x+7 = 0
Solving these two equations, we get
x = 1, x = \(\frac { 7 }{ 6 } \), x = \(\frac { 13 + \sqrt { 145 } }{ 12 } \) and x = \(\frac { 13-\sqrt { 145 } }{ 12 } \)
as the roots of the given equation.

We can solve this fourth degree equation by rewriting it suitably and adopting a technique of substitution. Rewriting the equation as
(x−2)(x−3)(x−7)(x+2)+19 = 0
the given equation becomes
(x²−5x+6)(x²−5x−14)+19 = 0 .
If we take x² − 5x as y, then the equation becomes (y+6)(y−14)+19 = 0;
that is, y²-8y-65 = 0
Solving this we get solutions y = 13 and y = −5. Substituting this we get two quadratic equations
x²-5x-13 = 0 and x²-5x+5 = 0
which can be solved by usual techniques. The solutions obtained for these two equations together give solutions as \(\frac { 5\pm \sqrt { 77 } }{ 2 } ,\frac { 5\pm \sqrt { 5 } }{ 2 } \).

Let the roots be in H.P. Then, their reciprocals are in A.P. and roots of the equation
\(\left( \frac { 1 }{ x } \right) ^{ 3 }+p\left( \frac { 1 }{ x } \right) ^{ 2 }+q\left( \frac { 1 }{ x } \right) \)+ r = 0 ⇔ rx³ + qx² + px + 1 = 0.....(1)
Since the roots of (1) are in A.P., we can assume them as α-d, α, α+d
Applying the Vieta’s formula, we get
Σ₁ = (α-d)+α+(α+d) = -\(\frac { q }{ r } \) ⇒ 3α = -\(\frac { q }{ r } \) ⇒ α = -\(\frac { q }{ 3r } \)
But, we note that α is a root of (1). Therefore, we get
\(r\left( -\frac { q }{ 3r } \right) ^{ 2 }+q\left( -\frac { q }{ 3r } \right) ^{ 2 }+p\left( -\frac { q }{ 3r } \right) \) + 1 = 0 ⇒ q³ + 3q³ - 9pqr + 27r² = 0 ⇒ 2q³ + 27r².