Let-a-b-c-p-be-rational-numbers-such-that-p-is-not-a-perfect-cube-If-a-bp-1-3-cp-2-3-0-then-prove-that-a-b-c-0-

Question Number 1083 by Vishal last updated on 08/Jun/15

L e t a, b, c, p b e r a t i o n a l n u m b e r s s u c h t h a t p i s n o t a p e r f e c t c u b e .

I f a + {b p}^{\frac{1}{3}} + {c p}^{\frac{2}{3}} = 0, t h e n p r o v e t h a t a = b = c = 0 .

Answered by prakash jain last updated on 08/Jun/15

p^{1 / 3} = x

Let c \neq 0

{c x}^{2} + b x + a = 0

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 c}

x = \frac{l}{m} + \frac{j}{k} \sqrt{n}

p i s n o t a p e r f e c t c u b e \Rightarrow n i s n o t a p e r f e c t s q u a r e .

\Rightarrow p = x^{3} i s i r r a t i o n a l c o n t r a c d i c t s t h a t p i s r a t i o n a l

h e n c e c = 0

∴ b x + a = 0

I f b \neq 0 x = \frac{- a}{b} \Rightarrow p = - \frac{a^{3}}{b^{3}} c o n t r a d i c t s t h e c o n d i t i o n

t h a t p i s n o t a p e r f e c t c u b e h e n c e b = 0

∵ b = 0, c = 0 \Rightarrow a = 0