Cubic equation...

Question

I did all I could to solve the question. I tried manipulating both the cubic equations to get the desired expression, I started out by assuming a function corresponding to these cubic equations and then doing manipulations but nothing seems to help.

Common values don't satisfy these cubic equations so surely they don't want us to solve the cubic. What I do realize is that the coefficients are well adjusted, so the solution definitely has got to do something with that.

Could you please help me with this?

Answer 1

Hint:

$a^3 -3a^2 +5a -4 = a^3+3a^2+5a +5 - 6a^2 -9 = 0 \implies 6a^2 = -9$. Here, we find the two complex solutions for the first polynomial. Find the real solution is easy from here.

Similiarly, $b^3 +3b^2 +5b +5 = b^3 -3b^2 +5b -4 +6b^2 + 9 = 0 \implies 6b^2 = -9$...

Answer 2

$$a^3-3a^2+5a-4=(a-1)^3+2(a-1)-1$$

$$b^3+3b^2+5b+5=(b+1)^3+2(b+1)+2$$

Considering $$f(x)=x^3+2x=x(x^2+2)$$ where $x=0$ is the only one real root and also the point of inflection (i.e. the centre of symmetry), thus $f(x)$ is increasing.

\begin{array}{|c|c|c|c|c|} \hline x & -1 & -\frac{ 1}{2} & 0 & \frac{ 1}{2} \\ \hline x^3+2x & -5 & -\frac{11}{8} & 0 & \frac{11}{8} \\ x^3+2x-1 & -6 & -\frac{19}{8} & -1 & \frac{ 3}{8} \\ x^3+2x+2 & -3 & \frac{ 5}{8} & 2 & \frac{27}{8} \\ \hline \end{array}

$$f(a-1)=1 \implies 0<a-1<\frac{1}{2}$$

$$f(b+1)=-2 \implies -1<b+1<-\frac{1}{2}$$

$$-1<a+b<0 \implies 0<|a+b|<1$$

Hence, $$[|a+b|]=0$$