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c,k be rational numbers such that
k is not a perfec
b^(2//3)+ck^(2//3) then prove that
a=b=c=0.

$c, k$ be rational numbers such that $k$ is not a perfec $b^{2 / 3}+c k^{2 / 3}$ then prove that $a=b=c=0$ .

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Write equations of cosine functions using properties

Full solution

Q.

c, k

be rational numbers such that

k

is not a perfec

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

then prove that

a=b=c=0

.

Cube Both Sides: Assume $a$ , $b$ , $c$ , and $k$ are rational numbers and $k$ is not a perfect cube. Given that $a^{\frac{2}{3}} + b^{\frac{2}{3}} + ck^{\frac{2}{3}} = 0$ , we can cube both sides to get rid of the fractional exponents. $\newline$ $(a^{\frac{2}{3}} + b^{\frac{2}{3}} + ck^{\frac{2}{3}})^3 = 0^3$
Expand Using Binomial Theorem: Expanding the left side using the binomial theorem, we get: $a^2 + b^2 + c^2k^2 + 3a^{\frac{2}{3}}b^{\frac{2}{3}} + 3a^{\frac{2}{3}}ck^{\frac{2}{3}} + 3b^{\frac{2}{3}}ck^{\frac{2}{3}} = 0$

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