28. f (2) - 4". then find the value of x3.
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x=3+2
2
1)
x
1
=
3+2
2
1
=
(3+2
2
)(3−2
2
)
3−2
2
=
9−8
3−2
2
=
1
3−2
2
(2)
x
3
−
x
3
1
=(x−
x
1
)(x
2
+
x
2
1
+x
x
1
)
x
3
−
x
3
1
=(x−
x
1
)(x
2
+
x
2
1
+1) (3)
x
2
+
x
2
1
=(x−
x
1
)
2
+2x
x
1
=(x−
x
1
)
2
+2 (4)
Substituting (4) in (3)
x
3
−
x
3
1
=(x−
x
1
)((x−
x
1
)
2
+2+1) (5)
Substituting (1) and (2) in (5)
x
3
−
x
3
1
=(3+2
2
−3+2
2
)((3+2
2
−3+2
2
)
2
+3)
x
3
−
x
3
1
=(4
2
)(35)=140
2
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