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 proove a=b=c=0

Question

let a b c p be rational numbers such that p is not a perfect
cube
if a+bp1/3+cp2/3=0
then proove a=b=c=0

Gursheen kaur. · Accepted Answer

$Given: a + b p^{\frac{1}{3}} + c p^{\frac{2}{3}} = 0 ... (1) \Rightarrow p^{\frac{1}{3}} (a + b p^{\frac{1}{3}} + c p^{\frac{2}{3}}) = 0 \Rightarrow a p^{\frac{1}{3}} + b p^{\frac{2}{3}} + c p = 0 ... (2)$

now, on multiplying (1) by b and (2) by c and subtracting, we get,

$(a b + b^{2} p^{\frac{1}{3}} + b c p^{\frac{1}{3}}) - (a c p^{\frac{1}{3}} + b c p^{\frac{2}{3}} + c^{2} p) = 0 \Rightarrow (b^{2} - a c) p^{\frac{1}{3}} + a b - c^{2} p = 0 \Rightarrow b^{2} - a c = 0 and a b - c^{2} p = 0 (∵ p^{\frac{1}{3}} is irrational) \Rightarrow b^{2} = a c and {a b = c}^{2} p ... (3) \Rightarrow b^{2} = a c and {a^{2} b^{2} = c}^{4} p^{2} \Rightarrow a^{2} (a c) = c^{2} p^{2} [from (3)] \Rightarrow a^{3} c - p^{2} c^{4} = 0 \Rightarrow a (a^{3} - p^{2} c^{3}) = 0 \Rightarrow c = 0 or a^{3} - p^{2} c^{3} = 0 \Rightarrow p^{2} = \frac{a^{3}}{c^{3}} \Rightarrow p^{\frac{2}{3}} = \frac{a}{c}$

but $p^{\frac{1}{3}}$ is irrational and $\frac{a}{c}$ is rational

$∴ a^{3} - p^{2} c^{3} \neq 0$

So, c = 0

$\Rightarrow b = 0 [f r o m (3)] \Rightarrow a = 0 [f r o m (1)]$

Hence a = b = c = 0

let a b c p be rational numbers such that p is not a perfect cube. if a+bp1/3+cp2/3=0 then proove a=b=c=0

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 proove a=b=c=0