<s>[INST] Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ 
Problem:Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ where $a,b,c$ are distinct real numbers
Solution:$(x-a)^3+(x-b)^3+(x-c)^3=0$
$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$
By Descartes rule of sign,number of positive real roots $=3$

But are they distinct $?$
Answer :- number of distinct real roots $ =1$
 [/INST] Let's substitute the variable: $x=y+\frac{(a+b+c)}{3}$
The equation will now look like 
$$3y^3+(2a^2-2ab-2ac+2b^2-2bc+2c^2)y+\frac{(a+b-2c)(a-3b+c)(b-2a+c)}{9}=0$$
Now we apply Cardano's method.
$$Q=\left(\frac{2a^2-2ab-2ac+2b^2-2bc+2c^2}{3}\right)^3+\left(\frac{\frac{(a+b-2c)(a-3b+c)(b-2a+c)}{9}}{2}\right)^2=\frac{8}{27}(a^2-ab-ac+b^2-bc+c^2)^3+\frac{1}{324}(a+b-2c)^2(a-2b+c)^2(b-2a+c)^2$$
Then the answer is given depending on the sign of Q.
If $Q>0$ there is only one real root
If $Q<0$ there are three distinct real roots
If $Q=0$ there are two distinct real roots.
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