Find the equation each of whose roots is greater by unity than a root of the equation $x^{3} - 5 x^{2} + 6 x - 3 = 0$ .

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Solution

$f (x) = x^{3} - 5 x^{2} + 6 x - 3 = 0$

Roots of the new equation are greater by 1, therefore $y = x + 1 ⟹ x = y - 1$

$⟹ f (y) = (y - 1)^{3} - 5 (y - 1)^{2} + 6 (y - 1) - 3 = 0$
$⟹ f (y) = y^{3} - 8 y^{2} + 19 y - 15 = 0$

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