Solve this equation: $7x^3+11x^2+6x+2=(x+2)\sqrt[3]{x^3+3x^2+5x+1}$

Question

Solve this equation: $7x^3+11x^2+6x+2=(x+2)\sqrt[3]{x^3+3x^2+5x+1}$

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I used Wolframalpha.com and get solutions $x\in\left\{0; \dfrac{-9\pm\sqrt{53}}{14}\right\}$. But I can't solve this.

Answer 1

Put $t = \sqrt[3]{x^3+3x^2+5x+1}.$ We have system of equations $$\begin{cases} 7 x^3 + 11 x^2 + 6 x + 2=(x+2)t,\\ t^3 = x^3+3x^2+5x+1. \end{cases}$$ Minus two equations, we get $$(-1 + t - 2 x)(4 x^2+(5 + 2 t) x+3 + t + t^2)=0.$$

The equation $$4 x^2+(5 + 2 t) x+3 + t + t^2=0$$ with unknown $x$ has discriminant is $-12 t^2+4 t-23$ negative for all $t$, therefore has no sulution.

The equation $-1 + t - 2 x = 0$ or $$\sqrt[3]{x^3+3 x^2+5 x+1}=2x+1$$ has three solutions $$x=0\lor x=\frac{1}{14}\left(-9-\sqrt{53}\right)\lor x=\frac{1}{14}\left(\sqrt{53}-9\right).$$

Answer 2

$7x^3+11x^2+6x+2 = (x+1)(7x^2+4x+2)$

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

Raising both sides to the $3$rd power:

$$(x+1)^3(7x^2+4x+2)^3=(x+2)^3((x+1)^3+2x) \implies \\ (x+1)^3(7x^2+4x+2)^3 - (x+2)^3(x+1)^3 - 2x(x+2)^3=0 \implies \\ (x+1)^3((7x^2+4x+2)^3-(x+2)^3) -2x(x+2)^3 = 0 \implies \\ x(x+1)^3(7x+3)(49x^4+63x^3+63x^2+30x+12) - 2x(x+2)^3 = 0 \implies \\ x (7 x^2+9 x+1) (49 x^6+168 x^5+266 x^4+262 x^3+175 x^2+78 x+20) = 0 $$

$x = 0$ is a solution, and $7x^2+9x+1 = 0$ is a solution. Compute the discriminant and find the roots.

Let $f(x) = 49 x^6+168 x^5+266 x^4+262 x^3+175 x^2+78 x+20$

$f'(x) = 294 x^5+840 x^4+1064 x^3+786 x^2+350 x+78$

$f''(x) = 2 (175 + 786 x + 1596 x^2 + 1680 x^3 + 735 x^4)$

$f'''(x) = 12 (131 + 532 x + 840 x^2 + 490 x^3)$

$f''''(x) = 168 (38 + 120 x + 105 x^2)$

$f''''$ is positive everywhere (compute discriminant, find that it is negative, etc.). So $f'''$ is increasing. Note that $f'''$ has exactly one root call it $\alpha$. Convince yourself that $f''(\alpha) > 0$ to deduce that $f'' > 0$ everywhere (or prove it, it is a good exercise). Continue until you reach the fact that $f > 0$ everywhere, so it contributes no solutions.

Answer 3

When disappear the radical the resulting equation is of degree 9 then, because C is algebraically closed, there are 9 roots (with eventual multiplicity). We know already 3 given by “idiots” and I admit my inability to calculate for myself directly the other 6 corresponding to the irreducible polynomial given by BolzWeir here. By computation they are with approximation (a, b) = (-1.0243, ±0.2351√(-1) ), (-0.1147, ±0.7073√(-1) ), (-0.5753, ±0.6236√(-1)).