If a+ bK^1/3+ cK^2/3=0, then prove …

We have equation 
{tex}a + b p ^ { \frac { 1 } { 3 } } + c p ^ { \frac { 2 } { 3 } } = 0{/tex} ...(i)
Multiplying both sides by {tex}p ^ { \frac { 1 } { 3 } }{/tex}in eq(i), we get
{tex}a p ^ { \frac { 1 } { 3 } } + b p ^ { \frac { 2 } { 3 } } + c p = 0{/tex} ...(ii)
Multiplying (i) by b and (ii) by c, we get
{tex}ab+b^2p^\frac13+bcp^\frac23=0{/tex} .......(iii) 

{tex}acp^\frac13+bcp^\frac23+c^2p=0{/tex} .......(iv).

subtracting equation (iv) from equation (iii) we get 

{tex}ab+b^2p^\frac13+bcp^\frac23-acp^\frac13-bcp^\frac23-c^2p=0{/tex}  

{tex}\Rightarrow\left(b^2p^\frac13-acp^\frac13\right)+\left(ab-c^2p\right)=0{/tex}
{tex}\Rightarrow \quad \left( b ^ { 2 } - a c \right) p ^ \frac{1}{3} + a b - c ^ { 2 } p = 0{/tex}
{tex}\Rightarrow \quad b ^ { 2 } - a c = 0 \text { and } a b - c ^ { 2 } p = 0 \quad \left[ \because p ^ { 1 / 3 } \text { is irrational } \right]{/tex}
{tex}\Rightarrow \quad b ^ { 2 } = a c \text { and } a b = c ^ { 2 } p{/tex}
{tex}\Rightarrow \quad b ^ { 2 } = a c \text { and } a ^ { 2 } b ^ { 2 } = c ^ { 4 } p ^ { 2 }{/tex}
{tex}\Rightarrow \quad a ^ { 2 } ( a c ) = c ^ { 4 } p ^ { 2 }{/tex} [Putting b² = ac in a²b² = c⁴p²]
{tex}\Rightarrow \quad a ^ { 3 } c - p ^ { 2 } c ^ { 4 } = 0{/tex}
{tex}\Rightarrow \quad \left( a ^ { 3 } - p ^ { 2 } c ^ { 3 } \right) c = 0{/tex}
{tex}\Rightarrow \quad a ^ { 3 } - p ^ { 2 } c ^ { 3 } = 0 , \text { or } c = 0{/tex}
Now, {tex}a ^ { 3 } - p ^ { 2 } c ^ { 3 } = 0{/tex}
 {tex}\Rightarrow p^2=\left(\frac ab\right)^3{/tex} 

cube root both side we get 

{tex}\left(p^2\right)^\frac13=\frac{a}{b}{/tex}  

{tex}\left(p^\frac13\right)^2=\frac{a}{b}{/tex}

this is not possible as {tex}p ^ { 1 / 3 }{/tex} is irrational and {tex}\frac { a } { b }{/tex} is rational.
{tex}\therefore \quad a ^ { 3 } - p ^ { 2 } c ^ { 3 } \neq 0{/tex}
Hence, c = 0
Putting c = 0 in b² = ac = 0, we get b = 0
Putting b = 0 and c = 0 in equation (i) {tex}a + b p ^ { 1 / 3 } + c p ^ { 2 / 3 } = 0{/tex}   

a + 0 +0 = 0, we get a = 0
Hence, a = b = c = 0.