The sum of the cubes of the roots of the equation, x^3 - px^2 + qx - r=0 is(A) p^3 + 3pq - 3r (B) p^3 + 3pq + 3r (C) p^3 - 3pg - 3r (D) p^3 - 3pg + 3r
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Answer:
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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