I don't see how you would do it without doing *some* calculation, so I'm not sure what the question is getting at. You can do all the steps of the proof in your head though. The argument should be completely clear from an intuitive geometric standpoint, and can be proved...

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I don't see how you would do it without doing *some* calculation, so I'm not sure what the question is getting at. You can do all the steps of the proof in your head though. The argument should be completely clear from an intuitive geometric standpoint, and can be proved using calculus.

Note that if we let

`f(x)=x(x-1)(x-3)+x(x-1)(x-2)+`

`x(x-2)(x-3)+(x-1)(x-2)(x-3),`

then `f(0)<0,` `f(1)>0,` `f(2)<0,` and `f(3)>0.`

There are no breaks in the graph of `f,` so it has a zero between 0 and 1, between 1 and 2, and between 2 and 3. It's a third degree polynomial, so it can not have more than three roots.

Here's the graph.

The equation provided is:

x(x-1)(x-3)+x(x-1)(x-2)+x(x-2)(x-3)+(x-1)(x-2)(x-3)=0

=> x(x - 3)(x - 1 + x - 2) + (x - 1)(x - 2)(x + x - 3) = 0

=> x(x - 3)(2x - 3) + (x - 1)(x - 2)(2x - 3) = 0

=> (2x - 3)(x^2 - 3x + x^2 - 3x + 2) = 0

=> (2x - 3)(2x^2 - 6x + 2) = 0

The linear term 2x - 3 has one solution.

The quadratic term 2x^2 - 6x + 2 has D = 36 - 16 = 20 indicating that 2x^2 - 6x + 2 = 0 has two real solutions.

**This shows that x(x-1)(x-3)+x(x-1)(x-2)+x(x-2)(x-3)+(x-1)(x-2)(x-3)=0 has three real roots.**