If $$(x - 5)^3 + (x - 6)^3 + (x - 7)^3 = 3 (x - 5) (x - 6) (x - 7)$$, then what is the value of x ?

As per the given question,

$$(x - 5)^3 + (x - 6)^3 + (x - 7)^3 = 3 (x - 5) (x - 6) (x - 7)$$

Let (x-5)=a, (x-6)=b and (x-7)=c

Now, $$(x - 5)^3 + (x - 6)^3 + (x - 7)^3 = 3 (x - 5) (x - 6) (x - 7)$$

$$\Rightarrow a^3 + b^3 + c^3 = 3 a b c$$

We know that $$a^3 + b^3 + c^3- 3 a b c=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

Hence, $$[(x-5)+(x-6)+(x-7)][(x-5)^2+(x-6)^2+(x-7)-(x-5)(x-6)-(x-6)(x-7)-(x-7)(x-5)]=0$$

$$\Rightarrow [(3x-18)][(x-5)^2+(x-6)^2+(x-7)-(x-5)(x-6)-(x-6)(x-7)-(x-7)(x-5)]=0$$

So, $$3x-18=0$$

Hence $$x=6$$