By neglecting x^4 and higher powers of x, fin | vedclass

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(B) Given expression: $\sqrt[3]{x^2+64}-\sqrt[3]{x^2+27}$ We can rewrite this as: $4(1+\frac{x^2}{64})^{1/3} - 3(1+\frac{x^2}{27})^{1/3}$ Using the bi

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$1-\frac{7}{432} x^2$

Vedclass · Answer

(B) Given expression: $\sqrt[3]{x^2+64}-\sqrt[3]{x^2+27}$ We can rewrite this as: $4(1+\frac{x^2}{64})^{1/3} - 3(1+\frac{x^2}{27})^{1/3}$ Using the binomial expansion $(1+u)^n \approx 1+nu$ for small $u$: $= 4[1 + \frac{1}{3}(\frac{x^2}{64})] - 3[1 + \frac{1}{3}(\frac{x^2}{27})] + \dots$ $= 4[1 + \frac{x^2}{192}] - 3[1 + \frac{x^2}{81}] + \dots$ $= 4 + \frac{4x^2}{192} - 3 - \frac{3x^2}{81} + \dots$ $= 1 + \frac{x^2}{48} - \frac{x^2}{27} + \dots$ $= 1 + x^2(\frac{9-16}{432}) = 1 - \frac{7}{432}x^2$ Thus, the correct option is $B$.

By neglecting $x^4$ and higher powers of $x$, find the approximate value of $\sqrt[3]{x^2+64}-\sqrt[3]{x^2+27}$.

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