let a, b, c and P be rational numbers such that P is not a perfect cube if a + bp the power 1 upon 3 + Dp power 2/3 equal to zero ,then prove that a=b=c=0
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★ QUADRATIC RESOLUTIONS ★
It's given that ,
a , b , c and p are rational numbers
such that p is not a perfect cube.
Therefore , p¹/³ and p²/³ are not rational numbers.
Therefore , p is neither 1 nor 0.
p²/³ = p¹/³ * p¹/³ and so they are unequal.
LHS = a + b p¹/³ + c p²/³ = 0 ---eq. (1)
Given p is not a perfect cube. p is not 0 or 1.
Also p¹/³,p²/³are irrational.
Now , Multiplying eq.(1) by p^1/3 to get:
c p + a p¹/³ + b p²/³ = 0 ---eq. (2)
eq. (1) / eq. (2)
a/cp = b/a = c/b
therefore , a²=cpb ; b²=ac ; c=b²/a
Substitute the value of c in (1) to get:
a²+abp¹/³+b²p²/³ = 0
D = -3a²b² which is negative ,
So p^1/3 is imaginary.
It is a contradiction as p is a rational number.
Given quadratic isn't valid.
So a = b = c = 0.
It's given that ,
a , b , c and p are rational numbers
such that p is not a perfect cube.
Therefore , p¹/³ and p²/³ are not rational numbers.
Therefore , p is neither 1 nor 0.
p²/³ = p¹/³ * p¹/³ and so they are unequal.
LHS = a + b p¹/³ + c p²/³ = 0 ---eq. (1)
Given p is not a perfect cube. p is not 0 or 1.
Also p¹/³,p²/³are irrational.
Now , Multiplying eq.(1) by p^1/3 to get:
c p + a p¹/³ + b p²/³ = 0 ---eq. (2)
eq. (1) / eq. (2)
a/cp = b/a = c/b
therefore , a²=cpb ; b²=ac ; c=b²/a
Substitute the value of c in (1) to get:
a²+abp¹/³+b²p²/³ = 0
D = -3a²b² which is negative ,
So p^1/3 is imaginary.
It is a contradiction as p is a rational number.
Given quadratic isn't valid.
So a = b = c = 0.
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