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The R.H.S.of the question can be easily manipulated to give 1 as the answer.
First of all consider:
[2+5^(1/2)]^1/3= [5^(1/2) +1]/2 and ;
[2-5^(1/2)]^1/3 =[1-5^(1/2) ]/2
Adding the above two equations clearly gives R.H.S.of the question
I.e. x=1
Let
x
=
3
2
+
5
+
3
2
−
5
.
x
−
3
2
+
5
−
3
2
−
5
=
0
.
Now let
a
=
x
,
b
=
−
3
2
+
5
and
c
=
−
3
2
−
5
.
Since
a
+
b
+
c
=
0
;
a
3
+
b
3
+
c
3
=
3
a
b
c
.
x
3
−
2
−
5
−
2
+
5
=
3
×
x
3
(
2
+
5
)
(
2
−
5
)
=
−
3
x
.
Therefore,
x
3
+
3
x
−
4
=
0
.
The rational root of this equation is
1
.
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x x 3 = 3 2 + 5 + 3 2 − 5 = ( 2 + 5 ) + 3 ( 2 + 5 ) 3 2 ( 2 − 5 ) 3 1 + 3 ( 2 + 5 ) 3 1 ( 2 − 5 ) 3 2 + ( 2 − 5 ) = 4 + 3 ( 2 + 5 ) 3 1 ( 2 − 5 ) 3 1 [ ( 2 + 5 ) 3 1 + ( 2 − 5 ) 3 1 ] = 4 + 3 ( 4 − 5 ) 3 1 x = 4 − 3 x
⟹ x 3 + 3 x − 4 ( x − 1 ) ( x 2 + x + 4 ) ⟹ x = 0 = 0 = 1
x 2 + x + 4 has no real root.