Solving a cubic

Algebra Level pending

The two irrational solutions to the equation 3 i x ( 4 − 3 x 2 − 2 i x ) = 8 3ix(4-3x^2-2ix)=8 can be represented as ± a b c \pm \frac{a\sqrt{b}}{c} , where a , b , c a,b,c are positive integers, a , c a,c are relatively prime, and b b is not divisible by the square of any integer greater than 1 1 . Find the product a b c abc .


The answer is 18.

Daniel Liu by Daniel Liu , 21, USA

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1 solution

Expand and simplify to get 9 x 3 − 6 x 2 + 12 x − 8 = 0 9x^3-6x^2+12x-8=0 .Apply the rational root test to find that 2 3 \frac{2}{3} is a root,or that x − 2 3 → 3 x − 2 x-\frac{2}{3}\rightarrow 3x-2 is a factor.Factoring the equation we get: 9 x 3 − 6 x 2 + 12 x − 8 = 0 3 x 2 ( 3 x − 2 ) + 4 ( 3 x − 2 ) = 0 ( 3 x − 2 ) ( 3 x 2 + 4 ) = 0 9x^3-6x^2+12x-8=0\\3x^2(3x-2)+4(3x-2)=0 \\ (3x-2)(3x^2+4)=0 3 x 2 + 4 3x^2+4 has roots ± 2 i 3 → ± 2 3 i 3 \pm\frac{2i}{\sqrt{3}}\rightarrow \pm\frac{2\sqrt{3}i}{3} So a = 2 , b = 3 , c = 3 → a b c = ( 2 ) ( 3 ) ( 3 ) = 18 a=2,b=3,c=3\rightarrow abc=(2)(3)(3)=\boxed{18}

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