Is there an easier way to see this factorisation

Is there an easier way for me to group $(xy+yz+xz)(x+y+z)$ into $(x+y)(y+z)(x+z)+xyz$ without fully expanding the brackets? Because once I get to $x^2y+y^2x+xyz+xyz+y^2z+yz^2+zx^2+xyz+z^2x$ it's very hard for me to see the grouping of $(x+y)(y+z)(x+z)$ amongst all those terms. So I was wondering is there an easier way, like some terms that I can pull out of the bracket, so that I can group the above in a more logical manner?

You have a degree-$3$ homogeneous sum of $3^2=9$ terms, of which three are $xyz$. Ideally, we'd "factorize" this - up to an $xyz$ term outside the factorization, as per @JoséCarlosSantos's objection${}^\dagger$ - as a product of three linear factors. The full symmetry implies it's either $(x+y+z)^3$ (up to e.g. $xyz$ terms) or doesn't use powers of $x+y+z$ at all. Well, it can't be such a perfect cube, as that would be $27$ terms.

But the also fully symmetric $(x+y)(y+z)(z+x)$ is much more helpful; that's only $8$ terms. That's one too few, so gives us the conjecture $(xy+yz+zx)(x+y+z)=(x+y)(y+z)(z+x)+xyz$, as opposed (I suppose) to $(xy+yz+zx)(x+y+z)=(x+y+z)^3-19xyz$. The latter is clearly wrong; the $xyz$ coefficient is $3$ on the left, but $3!-19=-13$ on the right.

As for the other idea, $(x+y)(y+z)(z+x)$ has no perfect cubes, but it has two $xyz$ terms (partner $x,\,y,\,z$ or $y,\,z,\,x$) plus six others that aren't squarefree, namely $x^2y+xy^2+\cdots$. Meanwhile,$$(xy+yz+zx)(x+y+z)$$has the same composition, except it has three $xyz$ terms instead.

As for @HagenvonEitzen's point, you'll enjoy reading this.

${}^\dagger$ Terminology aside, I interpret your question as to how we can see$$(xy+yz+zx)(x+y+z)=(x+y)(y+z)(z+x)+xyz.$$

Here is one method. Write the elementary symmetric functions as $$ e_1 := x+y+z,\quad e_2 := xy+xz+yz,\quad e_3 := xyz. $$ They are the coefficients of the cubic $$ (t-x)(t-y)(t-z) = t^3 - e_1t^2 + e_2t - e_3. $$ In this equation substitute $\,e_1\,$ for $\,t\,$ to get $$ (e_1-x)(e_1-y)(e_1-z) = e_1^3 - e_1e_1^2 + e_2e_1 - e_3. $$ Notice that $$ e_1-x = y+z, \quad e_1-y = x+z,\quad e_1-z = x+y $$ and $\,e_1^3 - e_1e_1^2 = 0\,$ and use these to get $$ (y+z)(x+z)(x+y) = (xy+xz+yz)(x+y+z) - xyz. $$