If y = - left( {{x^3} + frac{{{x^6}}}{2} + fr | vedclass

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Question

(B) Given the series $y = - \left( {{x^3} + \frac{{{x^6}}}{2} + \frac{{{x^9}}}{3} + \dots} \right)$.We know the logarithmic expansion $\ln(1 - t) = -

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$(1 - {e^y})^{1/3}$

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(B) Given the series $y = - \left( {{x^3} + \frac{{{x^6}}}{2} + \frac{{{x^9}}}{3} + \dots} \right)$.We know the logarithmic expansion $\ln(1 - t) = - \left( {t + \frac{{{t^2}}}{2} + \frac{{{t^3}}}{3} + \dots} \right)$ for $|t| Substituting $t = x^3$,we get $y = \ln(1 - x^3)$.Taking the exponential of both sides,we have $e^y = 1 - x^3$.Rearranging for $x^3$,we get $x^3 = 1 - e^y$.Therefore,$x = (1 - e^y)^{1/3}$.

If $y = - \left( {{x^3} + \frac{{{x^6}}}{2} + \frac{{{x^9}}}{3} + \dots} \right)$,then $x = $

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