Question

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✴️ CLASS 10 CBSE MATHS ✴️


Let a, b, c, k be rational numbers such that k is not a perfect cube.
If
$a + b {k}^{ \frac{1}{3} } + c {k}^{ \frac{2}{3} } = 0$
Then, prove that a = b = c =0.

Answer 1

a+ bk1/3 + ck2/3 = 0 --------- 1

=> -ck2/3 = a + bk1/3
=> taking cubes
=> (a³ + b³k + c³k) + (3a²b)k1/3 + (3ab²)k2/3 = 0 ------- 2

from 1 and 2 we get 

b² = ac

b³k + c³k² = 2a²

hence at end we get :

a = 0 , b = 0 , c = 0 

therefore a = b = c = 0

hope it helped
i know how to solve this as once i was going through a same question and i learned how to do so

Answer 2

Heya,
Friend,

_____________________

a + bk 1/3 + ck 2/3 = 0
a + bk 1/3 = - ck 2/3

Cubing ,

a³ + b³k + 3 a² bk 1/3 + 3ab²k²/3 = c³k²
a³ + b³k + 3 a² bk 1/3 + 3ab²k²/3 + c³k² = 0

(a³+b³k + c³k² ) + (3a²b) (k1/3) + (3ab²) k²/3 =0

Comparing,

3 a b² = c
3(0)b² = c
--------------
0 = c
_____________________

3 a² b = b
a² = 0
----------------
a = 0
_____________________

a = a³ + b³ k + c³k²
a = 0 + b³ k + 0
b³k = 0
-------------------------
b = 0

_____________________

Hope this will help you :)