Question

Let a, b, c, k, be rational number such that k is not a perfect cube . if a+b k 1/3 + ck2/3=0 then prove that a=b=c=0

Answer 1

$a+b k^{\frac{1}{3}}+ck^{\frac{2}{3}}=0---(1)\\\\-c k^{\frac{2}{3}}=a+bk^{\frac{1}{3}}\\\\take\ cubes\\\\-c^3k^2=a^3+b^3k+3a^2bk^{\frac{1}{3}}+3ab^2k^{\frac{2}{3}}\\\\(a^3+b^3k+c^3k^2)+(3a^2b)k^{\frac{1}{3}}+(3ab^2)k^{\frac{2}{3}}=0 - ---(2)\\\\Compare\ coefficients\ to\ get:\\\\b^2=ac\\\\b^3k+c^3k^2=2a^2\\\\as\ b^2\ \textless \ 4ac,\ discriminant\ of\ (1)\ is\ negative.\\\\Hence,\ a=0,\ b=0,\ c=0,\ as\ no\ solution.$

Answer 2

Answer:

Step-by-step explanation:

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