In Arthur Engels "Problem Solving Strategies" book in the section on symmetric polynomials, he asks us to prove the identities below. I read up on expanding trinomials and got the quickest method to be a variation on Pascal's triangle. Is there a different method to prove these identities; perhaps recursively?
Thanks
$$x^2+y^2+z^2=x^2+y^2+z^2+2(xy+xz+yz)-2(xy+xz+yz)=$$ $$=(x+y+z)^2-2(xy+xz+yz)=\sigma_1^2-2\sigma_2;$$ $$x^3+y^3+z^3=\sum_{cyc}x^3=\sum_{cyc}(x^3+x^2y+x^2z)-\sum_{cyc}(x^2y+x^2z+xyz)+3xyz=$$ $$=\sum_{cyc}x^2(x+y+z)-\sum_{cyc}xy(x+y+z)+3xyz=$$ $$=(x^2+y^2+z^2)(x+y+z)-(xy+xz+yz)(x+y+z)+3xyz=$$ $$=(\sigma_1^2-2\sigma_2)\sigma_1-\sigma_2\sigma_1+3\sigma_3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3;$$ $$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=\sum_{cyc}(x^2y+x^2z)=$$ $$=\sum_{cyc}(x^2y+x^2z+xyz)-3xyz=\sum_{cyc}(x^2y+xy^2+xyz)-3xyz=$$ $$=\sum_{cyc}xy(x+y+z)-3xyz=(x+y+z)(xy+xz+yz)-3xyz=\sigma_1\sigma_2-3\sigma_3;$$ $$x^2y^2+x^2z^2+y^2z^2=\sum_{cyc}x^2y^2=\sum_{cyc}(x^2y^2+2x^2yz)-2\sum_{cyc}x^2yz=$$ $$=(xy+xz+yz)^2-2xyz(x+y+z)=\sigma_2^2-2\sigma_1\sigma_3$$ and $$x^4+y^4+z^4=\sum_{cyc}x^4=\sum_{cyc}(x^4+2x^2y^2)-2\sum_{cyc}x^2y^2=$$ $$=\left(\sum_{cyc}x^2\right)^2-2\sum_{cyc}x^2y^2=(\sigma_1^2-2\sigma_2)^2-2(\sigma_2^2-2\sigma_1\sigma_3)=$$ $$=\sigma_1^4-4\sigma_1^2\sigma_2+2\sigma_2^2+4\sigma_1\sigma_3.$$