If & and ß are the roots of the equation
Px3+ qx+r=0 then &3beta+beta3& is
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Answer:
∴∑α
3
=−(3r+p
3
−3pq)
Step-by-step explanation:
Put y=x
3
in the equation x
3
+r=−px
2
−qx
⇒y+r=−x(px+q)
⇒(y+r)
3
=−x
3
(px+q)
3
⇒(y+r)
3
=−y(p
3
x
3
+q
3
+3p
2
qx
2
+3pq
2
x)
⇒(y+r)
3
=−y(p
3
y+q
3
+3pq(px
2
+qx))
⇒(y+r)
3
=−y(p
3
y+q
3
+3pq(−x
3
−r))
⇒(y+r)
3
=−y(p
3
y+q
3
+3pq(−y−r))
⇒y
3
+(3r+p
3
−3pq)y
2
+(3r
2
+q
3
−3pqr)y+r
3
=0
Roots of this equation are α
3
,β
3
,γ
3
∴∑α
3
=−(3r+p
3
−3pq)
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