Question

sum of p of the square of q and the cube of r​

Answer 1

Denote the roots by a,b,c; then

a+b+c=p,bc+ca+ab=q.

Now a  

2

+b  

2

+c  

2

=(a+b+c)  

2

−2(bc+ca+ab)

=p  

2

−2q.

Again, substitute a,b,c for x in the given equation and add; thus

a  

3

+b  

3

+c  

3

−p(a  

2

+b  

2

+c  

2

)+q(a+b+c)−3r=0;

∴a  

3

+b  

3

+c  

3

=p(p  

2

−2q)−pq+3r

=p  

3

−3pq+3r.

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