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.

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.

Given polynomial equation is 
2x⁴+ 5x³−7x² + 8 = 0
Here a = 2, b = 5, c = -7, d = 0, e = 8
By Vieta's formula, 
\(\alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -5 }{ 2 } \)
\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { -7 }{ 2 } \)
\(\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =0\)
\(\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { 8 }{ 2 } =4\)
Given roots of quadratic equation are 
∝ + β + ૪ + \(\delta \) and ∝β૪\(\delta \)
∴ sum of the roots = (∝+β+૪+\(\delta \)) (∝β૪\(\delta \))
\(=\left( \frac { -5 }{ 2 } +4 \right) =\frac { -5+8 }{ 2 } =\frac { 3 }{ 2 } \)
\(=\left( \alpha +\beta +\gamma +\delta \right) (\alpha \beta \gamma \delta )\)
\(=\left( \frac { -5 }{ 2 } \right) (4)=\frac { -20 }{ 2 } =-10\)
∴ The required quadratic equation is x²-x 
(sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x\left( \frac { 3 }{ 2 } \right) -10=0\)
\(\Rightarrow { 2x }^{ 2 }-3x-20=0\)

Given roots are 1, 2 and 3
Here a = 1, β = 2 and ૪ = 3
A cubic polynomial equation whose roots are α, β, ૪ is
x³-(α+β+૪)+x²(αβ+β૪+૪α)x-∝β૪ = 0
⇒ x³-(1+1+2)x²(2+6+3)x-6 = 0
⇒ x³-6x²+11x-6 = 0