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Algebra Difficulty 4.9 AIME Find the answer

Let r,s,tr, s, t be the solutions to the equation x3+ax2+bx+c=0x^{3}+a x^{2}+b x+c=0. What is the value of (rs)2+(st)2+(rt)2(r s)^{2}+(s t)^{2}+(r t)^{2} in terms of a,ba, b, and c?c ?

A number or a short expression. Spacing and $ signs are ignored.

Solution

We have (x−r)(x−s)(x−t)=x3+ax2+bx+c(x-r)(x-s)(x-t)=x^{3}+a x^{2}+b x+c, so a=−(r+s+t),b=rs+st+rt,c=−rst.a=-(r+s+t), \quad b=r s+s t+r t, \quad c=-r s t . So we have (rs)2+(st)2+(rt)2=(rs+st+rt)2−2rst(r+s+t)=b2−2ac(r s)^{2}+(s t)^{2}+(r t)^{2}=(r s+s t+r t)^{2}-2 r s t(r+s+t)=b^{2}-2 a c

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