Let a,b,c,k be rational numbers such that k is not a perfect cube. If a+bk^(1//3)+ck^(2//3) then prove that a=b=c=0.

Eliminate cube root: Multiply the equation by k^(1/3) to eliminate the cube root. ak^(1/3) + bk^(2/3) + ck = 0System of equations: Now we have a system of two equations: 1) a + bk^(1/3) + ck^(2/3) = 0 2) ak^(1/3) + bk^(2/3) + ck = 0Subtract and simplify: Subtract the second equation from the first one multiplied by k^(1/3): (a - ak^(1/3)) + (bk^(1/3) - bk) + (ck^(2/3) - ck^(4/3)) = 0Distinct coefficients: Simplify the equation: a(1 - k^(1/3)) + b(k^(1/3) - k) + c(k^(2/3) - k^(4/3)) = 0Coefficients analysis: Since k is not a perfect cube, k^(1/3) is not an integer, and thus 1, k^(1/3), k^(2/3), and k are all distinct.Final coefficients analysis: Therefore, the coefficients of 1, k^(1/3), and k^(2/3) must all be zero for the equation to hold true: a(1 - k^(1/3)) = 0 b(k^(1/3) - k) = 0 c(k^(2/3) - k^(4/3)) = 0Since k^(1/3) is not 1, a must be 0. Since k^(1/3) is not k, b must be 0. Since k^(2/3) is not k^(4/3), c must be 0.

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

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Q. Let

a,b,c,k

be rational numbers such that

k

is not a perfect cube.

\newline

a+bk^{\frac{1}{3}}+ck^{\frac{2}{3}}

then prove that

a=b=c=0

Eliminate cube root: Multiply the equation by $k^{1/3}$ to eliminate the cube root. $\newline$ $ak^{1/3} + bk^{2/3} + ck = 0$
System of equations: Now we have a system of two equations: $\newline$ $1$ ) $a + bk^{\frac{1}{3}} + ck^{\frac{2}{3}} = 0$ $\newline$ $2$ ) $ak^{\frac{1}{3}} + bk^{\frac{2}{3}} + ck = 0$
Subtract and simplify: Subtract the second equation from the first one multiplied by $k^{1/3}$ : $\newline$ $(a - ak^{1/3}) + (bk^{1/3} - bk) + (ck^{2/3} - ck^{4/3}) = 0$
Distinct coefficients: Simplify the equation: $a(1 - k^{\frac{1}{3}}) + b(k^{\frac{1}{3}} - k) + c(k^{\frac{2}{3}} - k^{\frac{4}{3}}) = 0$
Coefficients analysis: Since $k$ is not a perfect cube, $k^{1/3}$ is not an integer, and thus $1$ , $k^{1/3}$ , $k^{2/3}$ , and $k$ are all distinct.
Final coefficients analysis: Therefore, the coefficients of $1$ , $k^{1/3}$ , and $k^{2/3}$ must all be zero for the equation to hold true: $\newline$ $a(1 - k^{1/3}) = 0$ $\newline$ $b(k^{1/3} - k) = 0$ $\newline$ $c(k^{2/3} - k^{4/3}) = 0$
Final coefficients analysis: Therefore, the coefficients of $1$ , $k^{1/3}$ , and $k^{2/3}$ must all be zero for the equation to hold true: $\newline$ $a(1 - k^{1/3}) = 0$ $\newline$ $b(k^{1/3} - k) = 0$ $\newline$ $c(k^{2/3} - k^{4/3}) = 0$ Since $k^{1/3}$ is not $1$ , $a$ must be $0$ . $\newline$ Since $k^{1/3}$ is not $k^{1/3}$ $1$ , $k^{1/3}$ $2$ must be $0$ . $\newline$ Since $k^{2/3}$ is not $k^{1/3}$ $5$ , $k^{1/3}$ $6$ must be $0$ .