if a and b are roots of equation x^2+x-7=0 then find a^3+b^3
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Answer:
⇒ α and β are the roots of the equation x
2
−7x+1=0
Here, a=1,b=−7,c=1
⇒ αβ=
a
c
=
1
1
=1 ----- ( 1 )
⇒ α
2
β
2
=1 ----- ( 2 )
⇒ α+β=
a
−b
=−
1
−7
=7 ----- ( 3 )
⇒ (α+β)
2
=α
2
+β
2
+2αβ
⇒ (7)
2
=α
2
+β
2
+2(1) [ Using ( 1 ) and ( 3 ) ]
⇒ 49=α
2
+β
2
+2
∴ α
2
+β
2
=47 ------- ( 4 )
Now,
⇒
(α−7)
2
1
+
(β−7)
2
1
=
(α−7)
2
(β−7)
2
(β−7)
2
+(α−7)
2
=
(β
2
−14β+49)(α
2
−14α+49)
β
2
−14β+49+α
2
−14α+49
=
α
2
β
2
−14αβ
2
+49β
2
−14α
2
β+196αβ−686β+49α
2
−686α+2401
α
2
+β
2
−14(α+β)+98
=
α
2
β
2
−14αβ(α+β)+49(α
2
+β
2
)+196αβ−686(α+β)+2401
47−14(7)+98
[ Using ( 3 ) and ( 4 ) ]
=
1−14(1)(7)+49(47)+196(1)−686(7)+2401
47
=
1−98+2303+196−4802+2401
47
=
1
47
∴
(α−7)
2
1
+
(β−7)
2
1
=47