Solve the irrational equation I can not solve the equation:$$\sqrt[4]{x^3-2x^2-5x+6}+\sqrt{x^2+5x+6}=0.$$ Can someone help me. Thanky veru much.
 A: Hint: 
$\sqrt A$ (and $\sqrt[4]A$), when defined, are non-negative. Hence the equation is equivalent to the system
$$\begin{cases}x^3-2x^2-5x+6=0,\\x^2+5x+6=0.
\end{cases}$$
A: First I solve the equation:$x^3-2x^2-5x+6=0.$ Becouse the sum of the coefficinet of the given equation is $0,$ $(1+(-2)+(-5)+6=0)$, one zero is 1. Now we have a divide $$(x^3-2x^2-5x+6):(x-1)=x^2-x-6$$
i.e.,
$$x^3-2x^2-5x+6=(x-1)(x^2-x-6)=(x-1)(x+2)(x-3),$$ becouse 
$$x^2-x-6=0\Rightarrow x_{1,2}=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-6)}}{2}=\frac{1\pm \sqrt{1+24}}{2}=\frac{1\pm5}{2}; x_1=-2;x_2=3.$$
Now we solve the equation: $x^2+5x+6=0\Rightarrow x_1=-3,x_2=-2\Rightarrow x^2+5x+6=(x+2)(x+3).$
Now we have:
$$\sqrt[4]{x^3-2x^2-5x+6}+\sqrt{x^2+5x+6}=0$$
$$\sqrt[4]{x^3-2x^2-5x+6}=-\sqrt{x^2+5x+6}/^4$$
$$x^3-2x^2-5x+6=(x^2+5x+6)^2$$
$$(x-1)(x+2)(x-3)=[(x+2)(x+3)]^2$$
$$(x-1)(x+2)(x-3)=(x+2)^2(x+3)^2$$
$$(x-1)(x+2)(x-3)-(x+2)^2(x+3)^2=0$$
$$(x+2)[(x-1)(x-3)-(x+2)(x+3)^2]=0$$
$$(x+2)(x^3+7x^2+25x+14)=0$$
$$x+2=0\Rightarrow x=-2$$
$$x^3+7x^2+25x+14=0$$
(you try to solve this equation, subs. $x=\frac{3y-7}{3},$ but will not earn real solution)
I hope you help
A: $$\sqrt[4]{x^3-2x^2-5x+6}=\sqrt[4]{(x+2)(x-1)(x-3)}\\\sqrt{x^2+5x+6}=\sqrt{(x+2)(x+3)}$$
 Both quantities must be zero then we get the system $$\begin{cases}(x+2)(x-1)(x-3)=0\\(x+2)(x+3)=0\end{cases}$$ It is apparent that $(x+2)=0\iff x=-2$ is the only solution.
