Exercise 1.1.1

Complete the demonstration (begun in the text) that the substitution x = y − b ∕ 3 transforms x 3 + b x 2 + cx + d into y 3 + py + q , where p and q are given by (1.2).

Answers

Proof. The equation A x 3 + B x 2 + Cx + D = 0 ( A , B , C , D ∈ ℂ , A ≠ 0 ) becomes, after division by A ,

x 3 + b x 2 + cx + d = 0 .

with b = B ∕ A , c = C ∕ A , d = D ∕ A . The substitution x = y − b 3 gives

x = y − b 3 , x 2 = y 2 − 2 b 3 y + b 2 9 , x 3 = y 3 − b y 2 + b 2 3 y − b 3 27 ,

so

x 3 + b x 2 + cx + d = ( y 3 − b y 2 + b 2 3 y − b 3 27 ) + b ( y 2 − 2 b 3 y + b 2 9 ) + c ( y − b 3 ) + d = y 3 + ( − b 2 3 + c ) y + ( 2 b 3 27 − bc 3 + d ) .

In conclusion, if x = y − b 3 , the equation x 3 + b x 2 + cx + d = 0 is equivalent to the equation

y 3 + py + q = 0 ,

where

p = − b 2 3 + c , q = 2 b 3 27 − bc 3 + d ,

which is the equation (1.2).

Relatively to the first coefficients A , B , C , D we obtain

p = − B 2 3 A 2 + C A = − B 2 + 3 AC 3 A 2 , q = 2 B 3 27 A 3 − BC 3 A 2 + D A = 2 B 3 − 9 ABC + 27 A 2 D 27 A 3 . □
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2022-07-19 00:00
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