$f(x,y,z) = 6(x^5+y^5+z^5)-5(x^2+y^2+z^2)(x^3+y^3+z^3)$ is a homogenous symmetric polynomial of degree $5$. According to the theory of
symmetric polynomials, it can be uniquely expressed in terms of the elementary symmetric polynomials $e_1,e_2,e_3$, where $$e_1 = x+y+z,\qquad e_2 = xy+yz+zx, \qquad e_3 = xyz.$$ To write $f(x,y,z)$ in that form, use
Newton's identities, which say that if $p_k = x^k+y^k+z^k$ then $$p_1=e_1,\\ p_2 = e_1^2 - e_2,\\p_3 = e_1^3 - 3e_1e_2 + 3e_3,\\ p_k= e_1p_{k-1} - e_2p_{k-2} + e_3p_{k-3}\quad (k>3).$$ Then $$\begin{aligned}p_5& = e_1p_4 - e_2p_3 + e_3p_2\\ &= e_1(e_1p_3-e_2p_2+e_3p_1) - e_2p_3 + e_3p_2\\ &= (e_1^2-e_2)p_3 + (e_3-e_1e_2)p_2 + e_1e_3p_1\\ &=(e_1^2-e_2)(e_1^3 - 3e_1e_2 + 3e_3) + (e_3-e_1e_2)(e_1^2 - 2e_2) + e_1^2e_3\\ &= e_1^5 - 5e_1^3e_2 + 5e_1^2e_3 + 5e_1e_2^2 - 5e_2e_3.\end{aligned}$$ Therefore $$6p_5 = 6e_1^5 - 30e_1^3e_2 + 30e_1^2e_3 + 30e_1e_2^2 - 30e_2e_3.\qquad(*)$$ Next, $p_2p_3 = (e_1^2 - e_2)(e_1^3 - 3e_1e_2 + 3e_3) = e_1^5 - 5e_1^3e_2 + 3e_1^2e_3 + 6e_1e_2^2 - 6e_2e_3$, and so $$5p_2p_3 = 5e_1^5 - 25e_1^3e_2 + 15e_1^2e_3 + 30e_1e_2^2 - 30e_2e_3.\qquad(**)$$ From (*) and (**) it follows that $$f(x,y,z) = 6p_5 - 5p_2p_3 = e_1^5 - 5e_1^3e_2 + 15e_1^2e_3 = e_1^2(e_1^3 - 5e_1e_2 + 15e_3).$$ In terms of $x,y,z$, that says that $$6(x^5+y^5+z^5)-5(x^2+y^2+z^2)(x^3+y^3+z^3) = (x+y+z)^2\bigl(x^3+y^3+z^3 - 2x^2(y+z) - 2y^2(x+z) - 2z^2(x+y) + 6xyz\bigr)$$ (as you can verify by multiplying out the brackets on both sides, if you are so inclined).