Let a, b, c, p be rational numbers such that p is not a perfect cube.
Answers
Answer:
We have,
a+bp
3
1
+cp
3
2
=0
Multiplying both sides by p
3
1
, we get
ap
3
1
+bp
3
2
+cp=0
Multiplying (i)by b and (ii) by c and subtracting, we get
⎝
⎜
⎜
⎛
ab+b
2
p
3
1
+bcp
3
2
⎠
⎟
⎟
⎞
−
⎝
⎜
⎜
⎛
acp
3
1
+bcp
3
2
+c
2
p
⎠
⎟
⎟
⎞
=0
⇒(b
2
−ac)p
3
1
+ab−c
2
p=0
⇒b
2
−ac=0 and ab−c
2
p=0
⇒b
2
=ac and ab=c
2
p
⇒b
2
=ac and a
2
b
2
=c
4
p
2
⇒a
2
(ac)=c
4
p
2
⇒a
3
c−p
2
c
4
=0
⇒(a
3
−p
2
c
3
)c=0
⇒a
3
−p
2
c
3
=0 or c=0
Now,
a
3
−p
2
c
3
=0
p
2
=
c
3
a
3
⇒(p
2
)
3
1
=(
c
3
a
3
)
3
1
⇒
⎝
⎜
⎜
⎛
p
3
1
⎠
⎟
⎟
⎞
2
={(
c
a
)
3
}
3
1
⇒
⎝
⎜
⎜
⎛
p
3
1
⎠
⎟
⎟
⎞
2
=
c
a
This is not possible as p
1/3
is irrational and
c
a
is rational.
∴a
3
−p
2
c
3
=0
Hence c=0
Putting c=0 in b
2
−ac=0, we get b=0
Putting b=0 and c=0 in a+bp
3
1
+cp
3
2
=0 we get a=0
Hence a=b=c=0
Step-by-step explanation:
hope it help