The approximate value of (7.995)^{1/3} correc | vedclass

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(B) We can write $(7.995)^{1/3}$ as $(8 - 0.005)^{1/3}$.Using the binomial expansion $(1 - x)^n \approx 1 - nx + \frac{n(n-1)}{2!}x^2$ for small $x$:$

Vedclass · Accepted Answer

$1.9996$

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(B) We can write $(7.995)^{1/3}$ as $(8 - 0.005)^{1/3}$.Using the binomial expansion $(1 - x)^n \approx 1 - nx + \frac{n(n-1)}{2!}x^2$ for small $x$:$(8 - 0.005)^{1/3} = 8^{1/3} \left(1 - \frac{0.005}{8}ight)^{1/3} = 2 \left(1 - \frac{0.005}{8}ight)^{1/3}$.Applying the expansion $(1 - x)^n \approx 1 - nx$:$2 \left(1 - \frac{1}{3} 	imes \frac{0.005}{8}ight) = 2 \left(1 - \frac{0.005}{24}ight)$.$2 \left(1 - 0.00020833...ight) = 2(0.99979166...)$.$= 1.9995833...$.Rounding to four decimal places,we get $1.9996$.

The approximate value of $(7.995)^{1/3}$ correct to four decimal places is

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