<s>[INST] If $x^2-3x+9=0$, can we say $(x+3)(x^2-3x+9)$ is also $0$ hence $x^3=-27$? I was solving this question 

If $\dfrac x3 + \dfrac 3x = 1$ then find the value of $x^3$.

I solved it as. Cube both sides and substitute $x^3$ with $t$, 
$$ \dfrac{t}{27} + \dfrac{27}{t}=-2$$
$$\implies t^2 + {27}^2 + 2\cdot 27 \cdot t = 0$$
$$\implies (t+27)^2=0$$
$$\implies t=x^3=-27$$
Someone over facebook put another solution as,
$$\dfrac x3 + \dfrac 3x = 1$$
$$\implies x^2-3x+9=0$$
Multiplying both the sides with $(x+3)$,  
$$(x+3)(x^2-3x+9)=0$$
$$\implies x^3+27=0$$
Hence $x^3=-27$.
My question is, is it correct to multiply both sides with $0$? ( $x+3$ is $0$ here because $x^3=-27 \implies x=-3$ is one possible solution and other two being complex numbers.)
 [/INST] Yes there is nothing as such wrong with this process since it really helps in identifying the roots.
BUT one thing to keep in mind is that when you multiply the expression with $(x+3)$, you are introducing extraneous roots. That is, you had a quadratic equation with 2 roots and you have now made it cubic with 3 roots. So after solving, you must carefully eradicate this extraneous root you had introduced.
In the given problem, $x=-3$ is the extraneous root.
However you could have approached the problem using the discriminant.Then you get the roots easily.
$$x^2-3x+9=0 \Rightarrow x=\frac{3 \pm \sqrt{9-36}}{2}=\frac{3 \pm 3\sqrt{3}i}{2}$$
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