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, \(\frac { -3+\sqrt { 5 } }{ 2 } ,\frac { -3-\sqrt { 5 } }{ 2 } \).

12x + 8x = 29x²- 4
This equation can be re-written as
12x³ - 29x² + 8x + 4 = 0

∴ x = 2 is a root and the remaining factor is 
12x²- 5x - 2
⇒ (3x-2)(4x+1) = 0
⇒ 3x-2 = 0 or 4x+1 = 0

⇒ x = \(\frac{2}{3}\)
x = \(\frac{-1}{4}\)
∴ The roots are 2, \(\frac{2}{3}\), \(\frac{-1}{4}\)

Let f(x) = 8x³- 2x² - 7x +3 = 0
Here sum of the co-efficients of odd terms = 8 - 7 = 1
and sum of the co-efficients of even terms = -2 +3 = 1
Hence, x = - 1 is a root of f(x)
Let us divide f(x) by (x + 1)

∴ The other factor is 8x² - 10x + 3
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-4(8)(3) } }{ 2\times 8 } \)
\(\Rightarrow x=\frac { 10\pm \sqrt { 100-96 } }{ 16 } \Rightarrow x=\frac { 10\pm 2 }{ 16 } \)
\(\Rightarrow x=\frac { 12 }{ 16 } \ or \)
\(x=\frac { 8 }{ 16 } \Rightarrow x=\frac { 3 }{ 4 } ,\frac { 1 }{ 2 } \)
∴ The roots are -1, \(\frac{1}{2},\frac{3}{4}\)

4^x- 3(2^{x + 2}) + 2⁵ = 0
\(\Rightarrow \left( { 2 }^{ 2 } \right) ^{ x }-3({ 2 }^{ x })({ 2 }^{ 2 })+{ 2 }^{ 5 }=0\)
\(\left( { 2 }^{ x } \right) ^{ 2 }-3\left( { 2 }^{ x } \right) ({ 2 }^{ 2 })=0\)
\(\left( { 2 }^{ x } \right) ^{ 2 }-12\left( { 2 }^{ x } \right) +32=0\)
Put \({ 2 }^{ x }=y\)
\({ y }^{ 2 }-12y+32=0\)
(y - 8)(y - 4) = 0
y = 8, 4 
Case (i) when \(y=8,{ 2 }^{ x }=8\Rightarrow { 2 }^{ x }={ 2 }^{ 3 }\Rightarrow x=3\)
Case (ii) when \(y=4,{ 2 }^{ x }=4\Rightarrow { 2 }^{ x }={ 2 }^{ 2 }x=\pm 2\)
∴ The roots are 2, 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, \(\frac{1}{2}\).

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, \(\frac{2}{3}\)

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.

We observe that the sum of the coefficients of the odd powers and that of the even powers are equal.
Hence −1 is a root of the equation.
To find other roots, we divide 2x³+11x²-9x-18 by x+1 and get 2x²+9x-18 as the quotient.
Solving this we get \(\frac{3}{2}\) and -6 as roots.
Thus -6, -1, \(\frac{3}{2}\) are the roots or solutions of the given equation.

The roots of x³+2x²+3x+4 = 0 are ∝, β, ૪
∴ ∝ + β + ૪ = -co-efficient of x² = -2   ...(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3   ...(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4      ...(3)
Form the equation whose roots are ∝-β-૪
∴ -∝-β-૪ = -(∝+β+૪)
= -(-2) = 2
∝β + β૪ + ૪∝ = 3
(-∝)(-β)(-૪) = -(∝β૪) = -(-4) = 4
∴ The required cubic equation is 
x³-(-∝-β-૪)x²+(∝β+β૪+૪∝)
x-[(-∝)(-β)(-૪)] = 0
⇒ x³-(2)x²+3x-4 = 0
⇒ x³-2x²+3x-4 = 0

The roots of x³+2x²+3x+4 = 0 are ∝, β, ૪
∴ ∝ + β + ૪ = -co-efficient of x² = -2   .........(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3   ..........(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4  .......(3)
From the cubic equation whose roots are \(\frac { 1 }{ \alpha } ,\frac { 1 }{ \beta } ,\frac { 1 }{ \gamma } \)
\(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } =\frac { \beta \gamma +\gamma \alpha +\alpha \beta }{ \alpha \beta \gamma } =\frac { 3 }{ -4 } =\frac { -3 }{ 4 } \)
\(\frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } =\frac { \gamma +\alpha +\beta }{ \alpha \beta \gamma } =\frac { -2 }{ -4 } =\frac { 1 }{ 2 } \)
\(\left( \frac { 1 }{ \alpha } \right) \left( \frac { 1 }{ \beta } \right) \left( \frac { 1 }{ \gamma } \right) =\frac { 1 }{ \alpha \beta \gamma } =\frac { 1 }{ -4 } =-\frac { 1 }{ 4 } \)
∴ The required cubic equation is 
\({ x }^{ 3 }-\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } \right) { x }^{ 2 }+\left( \frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } \right) x-\left( \frac { 1 }{ \alpha } ,\frac { 1 }{ \beta } ,\frac { 1 }{ \gamma } \right) \)
\(\Rightarrow { x }^{ 3 }+\frac { 3 }{ 4 } { x }^{ 2 }+\frac { 1 }{ 2 } x+\frac { 1 }{ 4 } =0\)
Multiplying by 4 we get, 
4x³ + 3x² + 2x + 1 = 0