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.