The sum of squares of roots of the equation $x^{\frac{2}{3}}+x^{\frac{1}{3}}-2=0$ is

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Aicharya · Accepted Answer

Given, equation is $ \begin{array}{rlrl} & & x^{2 / 3}+x^{1 / 3}-2 & =0 \ \Rightarrow & & x^{2 / 3}+x^{1 / 3} & =2 \ & \Rightarrow & \left(x^{2 / 3}+x^{1 / 3}ight)^3 & =2^3 \ & & & \ {\left[	herefore(a+b)^3ight.} & \left.=a^3+b^3+3 a b(a+b)ight] \end{array} $ $\begin{aligned} & \Rightarrow \quad x^2+x+3 x 	imes 2=8 \Rightarrow \ & x^2+7 x-8=0 \ & \Rightarrow \quad(x+8)(x-1)=0 \Rightarrow x=1 ;-8 \ & 	ext { Sum of square roots }=(1)^2+(-8)^2 \ & \quad=1+64=65\end{aligned}$

The sum of squares of roots of the equation $x^{\frac{2}{3}}+x^{\frac{1}{3}}-2=0$ is

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