Vtyjkyuk

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

$$x^2+y^2+z^2=x^2+y^2+z^2+2(xy+xz+yz)-2(xy+xz+yz)=$$
$$=(x+y+z)^2-2(xy+xz+yz)=sigma_1^2-2sigma_2;$$
$$x^3+y^3+z^3=sum_cycx^3=sum_cyc(x^3+x^2y+x^2z)-sum_cyc(x^2y+x^2z+xyz)+3xyz=$$
$$=sum_cycx^2(x+y+z)-sum_cycxy(x+y+z)+3xyz=$$
$$=(x^2+y^2+z^2)(x+y+z)-(xy+xz+yz)(x+y+z)+3xyz=$$
$$=(sigma_1^2-2sigma_2)sigma_1-sigma_2sigma_1+3sigma_3=sigma_1^3-3sigma_1sigma_2+3sigma_3;$$
$$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=sum_cyc(x^2y+x^2z)=$$
$$=sum_cyc(x^2y+x^2z+xyz)-3xyz=sum_cyc(x^2y+xy^2+xyz)-3xyz=$$
$$=sum_cycxy(x+y+z)-3xyz=(x+y+z)(xy+xz+yz)-3xyz=sigma_1sigma_2-3sigma_3;$$
$$x^2y^2+x^2z^2+y^2z^2=sum_cycx^2y^2=sum_cyc(x^2y^2+2x^2yz)-2sum_cycx^2yz=$$
$$=(xy+xz+yz)^2-2xyz(x+y+z)=sigma_2^2-2sigma_1sigma_3$$ and
$$x^4+y^4+z^4=sum_cycx^4=sum_cyc(x^4+2x^2y^2)-2sum_cycx^2y^2=$$
$$=left(sum_cycx^2right)^2-2sum_cycx^2y^2=(sigma_1^2-2sigma_2)^2-2(sigma_2^2-2sigma_1sigma_3)=$$
$$=sigma_1^4-4sigma_1^2sigma_2+2sigma_2^2+4sigma_1sigma_3.$$