x⁴+ 3x³- 3x - 1 = 0
Sum of the co-efficients = 1 + 3 - 3.- 1 = 0 
⇒x = 1 is a root ⇒(x - 1) is a factor

[Using synthetic division]
∴ x = 1, 1 are the roots and the remaining factor
\(\Rightarrow x=\frac { -3\pm \sqrt { 9-4(1)(1) } }{ 2 } \)
\(\Rightarrow x=\frac { -3\pm \sqrt { 5 } }{ 2 } \)
∴ The roots are 1, -1, \(\frac { -3+\sqrt { 5 } }{ 2 } ,\frac { -3-\sqrt { 5 } }{ 2 } \).

(x-4)(x-7)(x-2)(x+1) = 16
Given (x - 4)(x -7)(x - 2)(x + 1) = 16
Rearrange the terms as
(x - 4)(x - 2)(x -7)(x + 1) = 16
⇒ (x² - 6x + 8)(x2 - 6x -7) = 16
Put x²-6x = y
⇒ (y + 8) (y - 7) = 16
⇒ y² +y - 56 - 16 = 0
⇒ (y+9)(y-8) = 0
⇒ y = -9, 8

case (i)
 When y = -9
⇒ x²-6x = -9
⇒ x²-6x+9 = 0
⇒ (x-3)² = 0
⇒ x = 3, 3

case ii
When y = 8,
\(\Rightarrow x=\frac { 6\pm \sqrt { 36-4(1)(-8) } }{ 2 } \)
\(\Rightarrow x=\frac { 6\pm \sqrt { 36-32 } }{ 2 } \Rightarrow x=\frac { 6\pm \sqrt { 68 } }{ 2 } \)
\(\Rightarrow x=\frac { 6\pm 2\sqrt { 17 } }{ 2 } \Rightarrow x=\frac { 2(3\pm \sqrt { 17 } ) }{ 2 } \)
\(\Rightarrow x=3\pm \sqrt { 17 } \)
∴ The roots are 3, 3, 3 +\(\sqrt{17}\), and 3 -\(\sqrt{17}\)

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 \(2,\frac { 1 }{ 2 } ,3,\frac { 1 }{ 3 } \)

This equation is Type I even degree reciprocal equation. Hence it can be rewritten as
x²\(\left[ \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -10 \left( x+\frac { 1 }{ x } \right) +26 \right] \)= 0 Since x \(\neq\) 0, we get \(\left(x^{2}+\frac{1}{x^{2}}\right)-10\left(x+\frac{1}{x}\right)+26=0\)
Let y = \(\left( x+\frac { 1 }{ x } \right) \)
[(y²-2)-10y+26] = 0 ⇒ (y²-10y+24) = 0 ⇒ (y-6)(y-4) = 0 ⇒ y = 6 or y = 4
Case (i)
y = 6 ⇒ x +\(\frac{1}{x}\) = 6 ⇒ x = 3+2\(\sqrt{2}\), x = 3 - 2\(\sqrt{2}\)
Case (ii)
y = 4 ⇒ x = 2+\(\sqrt{3}\), x = 2-\(\sqrt{3}\).
Hence, the roots are \(3 \pm 2 \sqrt{2}, 2 \pm \sqrt{3}\)

Put x² = y
⇒ y²- 14y + 45 = 0
⇒ (y - 9) (y - 5) = 0
⇒ y = 9 or y = 5
⇒ x² = 9 or x² = 5
⇒ x = ±33 or x =±3√5
Hence the roots are 3, -3, √5 and -√5.

Let f(x) x⁶-3x⁵-5x⁴+22x³-39x²-39x+135
Given (1+2i) is a root \(\Rightarrow\)(-2i) is also a root
Also \(\sqrt3\) is a root \(\Rightarrow\)-\(\sqrt3\) is also a root.
Hence, the factors of f(x) are [x - (1 + 2i)]
[x-(1-2i)] [x\(\sqrt3\)] [x+\(\sqrt3\)]
[(x-1)-2i] [(x-1)+2i] [x-\(\sqrt3\)][x+\(\sqrt3\)]
((x-1)²+2²)(x²-3) = (x²-2x+1+4)(x²-3)
\(\Rightarrow\) factor of f(x) is (x²-2x+5)(x²-3)
\(\Rightarrow\)x⁴-3x²-2x³+6x+5x²-15
\(\Rightarrow\)(x⁴-3x²-2x³+6x-15) is a factor of f(x) 
To find the other factor, let us divide f(x) by
x⁴ - 2x³ + 2x² + 6x - 15

The other factor is x² - x - 9
\(\Rightarrow x=\frac { 1\pm \sqrt { { (-1) }^{ 2 }-4(1)(-9) } }{ 2 } \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
\(\Rightarrow x=\frac { 1\pm \sqrt { 37 } }{ 2 } \)
Hence the roots are
1 - 2i, 1 + 2i, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

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
\(\Rightarrow \frac { 4 }{ 3 } -\frac { 2 }{ 3 } ,\frac { 4 }{ 3 } ,\frac { 4 }{ 3 } +\frac { 2 }{ 3 } \Rightarrow \frac { 2 }{ 3 } ,\frac { 4 }{ 3 } ,2\)

Rearrange the terms as
(2x-1)(2x+ 3) (x + 3) (x - 2) + 20 = 0
⇒ (4x² + 6x - 2x- 3)(x² - 2x + 3x - 6) + 20 = 0
⇒ (4x² + 4x - 3) (x² + x - 6) + 20 = 0
put x²+ x = y
⇒ (4y - 3) (y - 6) + 20 = 0
⇒ 4y² - 24y - 3y + 18 + 20 = 0
⇒ 4y²-27y +38 = 0
⇒ (y - 2)( 4y - 19) = 0
\(y=2,\frac { 19 }{ 4 } \)

Case (i) 
When y = 2
x²+ x = 2
x² + x - 2 = 0
⇒ (x + 2)(x - 1) = 0
⇒ x = -2, 1
Case (ii) 
When \(y=\frac { 19 }{ 4 } ,{ x }^{ 2 }+x=\frac { 19 }{ 4 } \)
\(\Rightarrow { 4x }^{ 2 }+4x=19\)
\(\Rightarrow { 4x }^{ 2 }-4x-19=0\)
\(\Rightarrow x=\frac { -4\pm \sqrt { 16-4(4)(-19) } }{ 8 } \)
\(\Rightarrow x=\frac { -4\pm \sqrt { 16+304 } }{ 8 } \)
\(\Rightarrow x=\frac { -4\pm \sqrt { 320 } }{ 8 } \)
\(\Rightarrow x=\frac { 4\pm 8\sqrt { 5 } }{ 8 } \)
\(\Rightarrow x=\frac { -4(-1\pm 2\sqrt { 5 } ) }{ 8 } \)
\(\frac{-1 \pm 2 \sqrt{5}}{2}\)
Hence the roots are 1, -2, \(\frac{-1 \pm 2 \sqrt{5}}{2}\)

(i) 
Rearrange the terms as,
(x-5) (x+4) (x-7) (x+6) = 504
{-2, 3, -7, 8}
⇒ (x² -x - 20) (x² -x - 42) = 504
Put x²- x = y
⇒ (y-20)(y-42) = 504
⇒ y²-62y+840-504 = 0
⇒ y²-62y+336 = 0
⇒ (y - 56) (y - 6) = 0
⇒ y = 56, 6
Case (i)
When y = 56, x² - x = 56

⇒ x² - x - 56 = 0
⇒ (x - 8)(x + 7) = 0
⇒ x = 8, -7
Case (ii) 
When y = 6,
x²- x = 6
x²- x - 6 = 0
⇒ (x - 3)(x + 2) = 0
⇒ x = 3, - 2

Hence the roots are -2, 3, 8, -7

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 G.P.
Then, we can assume them in the form \(\frac { \alpha }{ \lambda } \), α, αλ.
Applying the Vieta’s formula, we get
Σ₁ = \(\alpha \left( \frac { 1 }{ \lambda } +1+\lambda \right) =\frac { b }{ a } \) ....(1)
Σ₂ = \(\alpha ^{ 2 }\left( \frac { 1 }{ \lambda } +1+\lambda \right) =\frac { c }{ a } \) ............(2)
Σ₃ = α³ = -\(\frac { d }{ a } \) ..........(3)
Dividing (2) by (1), we get
α = -\(\frac { c }{ b } \) ...........(4)
Substituting (4) in (3), we get \(\left( -\frac { c }{ b } \right) ^{ 3 }=\frac { d }{ a } \) ⇒ ac³ = db³.

Since the coefficients of the equation are all rational numbers, 2+i and 3-\(\sqrt{2}\) are roots, we get 2-i and 3+\(\sqrt{2}\) are also roots of the given equation. Thus (x-(2+i)), (x-(2-i)), (x-(3-\(\sqrt{2}\))) and (x-(3+\(\sqrt{2}\))) are factors. Thus their product.
((x-(2+i))(x-(2-i))(x-(3-\(\sqrt{2}\)))(x-(3+\(\sqrt{2}\))) is a factor of the given polynomial equation.
That is, (x²-4x+5)(x²-6x+7) is a factor. Dividing the given polynomial equation by this factor, we get the other factor as (x²-3x-4) which implies that 4 and −1 are the other two roots. Thus
2+i, 2-i, 3+\(\sqrt{2}\), 3-\(\sqrt{2}\), -1, and 4 are the roots of the given polynomial equation.

By choosing the co-ordinate axes suitably, we take the equation of the straight line as
y = mx + c .........(1)
and equation of parabola as y² = 4ax ............(2)
Substituting (1) in (2),we get
(mx + c)² = 4ax
⇒ m²x² + c² + 2mcx = 4ax
⇒ m²x²+x(2mc - 4a) + c² = 0
Which is a quadratic equation in x.
This equation cannot have more than two solution. Hence, a straight line and a parabola cannot intersect at more than two points.

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

Let ∝, β, ૪ be the roots of the equation 
Given \(\frac { \alpha }{ \beta } =\frac { 3 }{ 2 } \Rightarrow 2\alpha =3\beta \Rightarrow \alpha =\frac { 3 }{ 2 } \beta \)
\(\therefore \frac { 3 }{ 2 } \beta ,\beta ,\gamma \) are the roots of the given equation
Then by Vieta's formula,
\(\frac { 3 }{ 2 } \beta +\beta +\gamma =\frac { -b }{ a } =\frac { -(-9) }{ 1 } =9\)
\(\frac { 5 }{ 2 } \beta +\gamma =9\Rightarrow \gamma =9-\frac { 5 }{ 2 } \beta \)
\(\Rightarrow \gamma =\frac { 18-5\beta }{ 2 } ...(2)\)
Also \(\frac { 3 }{ 2 } \beta (\beta )+\beta \gamma +\left( \frac { 3 }{ 2 } \beta \right) \gamma =\frac { c }{ a } =\frac { 14 }{ 1 } =14\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 5 }{ 2 } \beta \left( \frac { 18-5\beta }{ 2 } \right) =14\ [using\ (2)]\)
\(\Rightarrow \frac { 3 }{ 2 } { \beta }^{ 2 }+\frac { 90\beta }{ 4 } -\frac { 25{ \beta }^{ 2 } }{ 4 } =14\)
Multiplying by \(4,6{ \beta }^{ 2 }+90{ \beta }-25{ \beta }^{ 2 }=56\)
\(19{ \beta }^{ 2 }-90{ \beta }+56=0\)
\(\Rightarrow ({ \beta }-4)(19{ \beta }-14)=0\)
\(\Rightarrow { \beta }=4\)
\({ \beta }=\frac { 14 }{ 19 } \)
When \(\beta\) = 4, the other roots are \(\frac { 3 }{ 2 } (4),4,\frac { 18-5 }{ 2 } (4)\)
\(\Rightarrow 6,4,-1\)

When \(\\ \beta =\frac { 14 }{ 19 } ,\) the other roots are \(\frac { 3 }{ 2 } \beta ,\beta \frac { 18-5\beta }{ 2 } [by(2)]\)
\(\Rightarrow \frac { 3 }{ 2 } \left( \frac { 14 }{ 19 } \right) ,\frac { 14 }{ 19 } ,\frac { 18-5\left( \frac { 14 }{ 19 } \right) }{ 2 } \Rightarrow \frac { 21 }{ 19 } ,\frac { 14 }{ 19 } ,\frac { 136 }{ 19 } \)

Let p(-x) = x⁹- 5x⁵ + 4x⁴ + 2x² + 1 = 0
P(x) has only one sign change. It has atmost one positive roots.
Also p(-x) = (-x)⁹ - 5(-x)⁵ + 4(-x)4 +2(-x)² +1 = 0
p(-x) = -x⁹+ 5x⁵+4x⁴+ 2x²+1 = 0
It has only one sign change.
It has atmost one negative roots.
Clearly 0 is not a root.
So maximum number of real roots is 3 and hence there are atleast six imaginary solutions.

Given \((\sqrt { 5 } -\sqrt { 3 } )\) is a root
Another root \(\Rightarrow \sqrt { 5 } +\sqrt { 3 } \) 
∴ Sum of the roots \(=\sqrt { 5 } -\sqrt { 3 } +\sqrt { 5 } +\sqrt { 3 } =2\sqrt { 5 } \)
Product of the roots 
\(=(\sqrt { 5 } -\sqrt { 3 } )(\sqrt { 5 } +\sqrt { 3 } )\)
\(=(\sqrt { 5 } )^{ 2 }-{ (\sqrt { 3 } ) }^{ 2 }=5-3=2\)
∴ One of the factor is x² -x (sum of the roots) + product of the roots 
\(\Rightarrow { x }^{ 2 }-2x\sqrt { 5 } +2\)
The other factor also will be \({ x }^{ 2 }-2x\sqrt { 5 } +2\)
\(({ x }^{ 2 }-2x\sqrt { 5 } +2)({ x }^{ 2 }+2x\sqrt { 5 } +2)=0\)
\(\Rightarrow ({ x }^{ 2 }+2-2\sqrt { 5 } x)({ x }^{ 2 }+2+2\sqrt { 5 } x)=0\)
\(\Rightarrow \left( { x }^{ 2 }+2 \right) ^{ 2 }-{ (2\sqrt { 5 } x) }^{ 2 }=0\)
\([\because (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 }]\)
\(\Rightarrow { x }^{ 4 }+{ 4x }^{ 2 }+4{ (5)x }^{ 2 }=0\)
\(\Rightarrow { x }^{ 4 }+{ 4x }^{ 2 }+4-{ 20x }^{ 2 }=0\)
\(\Rightarrow { x }^{ 4 }-{ 16x }^{ 2 }+4=0\) is a rational co-efficient polynomial equation.

Since 2-\(\sqrt{3}\)i is a root and the coefficients are rational numbers, 2+\(\sqrt{3}\)i  is also a root. A required polynomial equation is given by
x² −(Sum of the roots) x + Product of the roots = 0
and hence
x²- 4x +1 = 0 is a required equation.

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 ∝, 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, \(\sum { \frac { \alpha }{ \beta \gamma } } =\frac { \alpha }{ \beta \gamma } +\frac { \beta }{ \gamma \alpha } +\frac { \gamma }{ \alpha \beta } \)
\(=\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } }{ \alpha \beta \gamma } \)
\(=\frac { { (\alpha +\beta +\gamma ) }^{ 2 }-2(\alpha \beta +\beta \gamma +\gamma \alpha ) }{ \alpha \beta \gamma } \)
\(=\frac { { \left( -\frac { b }{ a } \right) }^{ 2 }-2\left( \frac { c }{ a } \right) }{ -\frac { d }{ a } } \)
\(=\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } }{ -\frac { d }{ a } } \Rightarrow \frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \times \frac { -a }{ d } \)
\(\therefore \frac { \sum { \alpha } }{ \beta \gamma } =-\frac { \left( { b }^{ 2 }-2ac \right) }{- ad } =\frac { 2ac-{ b }^{ 2 } }{ ad } \)