$4$ . Let $a, b, c, k$ be rational numbers such that $k$ is not a perfect cube. $\newline$ If $a+b k^{1 / 3}+c k^{2 / 3}$ then prove that $a=b=c=0$ . $\newline$ Sol. Given, $a+b k^{i s}+c k^{2 \beta}=0$ $\newline$ Multiplying both sides by $k^{1 / 3}$ , we have $\newline$ $a k^{1 / 3}+b k^{2 / 3}+c k=0 \text {. }$

Full solution

4

. Let

a, b, c, k

be rational numbers such that

k

is not a perfect cube.

\newline

a+b k^{1 / 3}+c k^{2 / 3}

then prove that

a=b=c=0

\newline

Sol. Given,

a+b k^{i s}+c k^{2 \beta}=0

\newline

Multiplying both sides by

k^{1 / 3}

, we have

\newline

a k^{1 / 3}+b k^{2 / 3}+c k=0 \text {. }

Multiply by $k^{1/3}$ : Multiply the equation by $k^{1/3}$ to eliminate the cube root. $ak^{1/3} + bk^{2/3} + ck = 0$
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 $k^{1/3}$ : $\newline$ $(a + bk^{1/3} + ck^{2/3})k^{1/3} - (ak^{1/3} + bk^{2/3} + ck) = 0$ $\newline$ This simplifies to: $\newline$ $ak^{2/3} + bk + ck^{4/3} - ak^{2/3} - bk^{4/3} - ck = 0$
Cancel Like Terms: Simplify the equation by canceling out like terms: $\newline$ $bk + ck^{\frac{4}{3}} - bk^{\frac{4}{3}} - ck = 0$ $\newline$ This simplifies to: $\newline$ $bk - bk^{\frac{4}{3}} + ck^{\frac{4}{3}} - ck = 0$
Factor Out Common Terms: Factor out the common terms: $b(k - k^{4/3}) + c(k^{4/3} - k) = 0$
Factors Not Zero: Since $k$ is not a perfect cube, $k^{1/3}$ is not an integer, and thus $k^{4/3}$ is not equal to $k$ . Therefore, the factors $(k - k^{4/3})$ and $(k^{4/3} - k)$ are not zero.
Solve for $b$ and $c$ : For the equation to hold true, since the factors are not zero, both $b$ and $c$ must be zero: $\newline$ $b = 0$ , $c = 0$
Substitute Back into Equation: Substitute $b = 0$ and $c = 0$ back into the original equation: $\newline$ $a + 0\cdot k^{1/3} + 0\cdot k^{2/3} = 0$ $\newline$ This simplifies to: $\newline$ $a = 0$