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Transform the following equation into an equation whose roots are four times those of given equation 3x- 2x- x + 1 = 0.

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Given equation is 

3x3 - 2x2 - x + 1 = 0----(1)

Let α, ß, γ  are root of equation (1)

\(\therefore\) Sum of roots = -b/a = -(-2)/3 = 2/3

⇒ α + ß + γ  = 2/3 ----(2)

sum of product of two roots = c/a = -1/3

⇒ αß + ßγ + γα = -1/3----(3)

⇒ αßγ = -1/3 ---(4)

Now, 4α + 4ß + 4γ = 4(α + ß + γ) = 4 x 2/3 = 8/3 (From 2)

4α .4ß + 4ß.4γ + 4γ. 4α = 16(αß + ßγ + γα)

 = 16 x -1/3 = -16/3 (From 3)

4α. 4ß. 4γ = 64αßγ = 64 x -1/3 = -64/3 (From (4))

\(\therefore\) Required equation whose roots are 4 times than given roots is x3 - (sum of roots)x2 + (sum of products of two roots)x - (products of roots) = 0

⇒ x3 - (8/3)x2 = (16/3) x - \((-64/3)=0\)

⇒ 3x3 - 8x2 - 16x + 64 = 0 which is required equation.

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