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Expand the quantity cube root {P + t} about 0 in terms of t / P. Give four nonzero terms.

Question:

Expand the quantity {eq}\displaystyle \sqrt[3] {P + t} {/eq} about 0 in terms of {eq}\displaystyle \frac{t }{P} {/eq}. Give four nonzero terms.

Binomial Expansion for Cube Root:

Firstly, we'll rewrite the cube root function as the fractional exponent function and then use the algebraic rule of the exponent for the common base.

  • {eq}\displaystyle \sqrt[3]{x}=x^{\frac{1}{3}}\\ (xy)^m=x^m\cdot y^m {/eq}

After that, we have an expression such as {eq}\displaystyle (1+x)^m {/eq} then we'll expand the obtained expression using the binomial formula.

  • {eq}\displaystyle (1+x)^m=1+mx+\frac{m(m-1)}{2!}x^2+\frac{m(m-1)(m-2)}{3!}x^3+\dots {/eq}

Answer and Explanation:

The given cube root function is:

{eq}\displaystyle \sqrt[3] {P + t} {/eq}

Rewriting the above cube root function as fractional exponent, we get:

{eq}\displaystyle \sqrt[3] {P + t}=(P+t)^{\frac{1}{3}} {/eq}


Taking a term out from the above expression, we get:

{eq}\begin{align*} \displaystyle (P+t)^{\frac{1}{3}}&=(P(1+\frac{t}{P}))^{\frac{1}{3}}\\ &=P^{\frac{1}{3}}(1+\frac{t}{P})^{\frac{1}{3}}&\because (xy)^m=x^m\cdot y^m\\ \end{align*} {/eq}


Here, we have:

{eq}x=\frac{t}{P}\\ m=\frac{1}{3} {/eq}

Expanding the expression {eq}(1+\frac{t}{P})^{\frac{1}{3}} {/eq} using the binomial formula, we get:

{eq}\begin{align*} \displaystyle (1+\frac{t}{P})^{\frac{1}{3}}&=1+\frac{1}{3}\left ( \frac{t}{P} \right )+\frac{\frac{1}{3}\left ( \frac{1}{3}-1 \right )}{2!}\left ( \frac{t}{P} \right )^2+\frac{\frac{1}{3}\left ( \frac{1}{3}-1 \right )\left ( \frac{1}{3}-2\right )}{3!}\left ( \frac{t}{P} \right )^3+\dots\\ &=1+\frac{t}{3P}+\frac{\frac{1}{3}\left ( -\frac{2}{3} \right )}{2(1)}\left ( \frac{t^2}{P^2} \right )+\frac{\frac{1}{3}\left (- \frac{2}{3} \right )\left ( -\frac{5}{3}\right )}{3(2)(1)}\left ( \frac{t^3}{P^3} \right )+\dots\\ &=1+\frac{t}{3P}-\frac{1}{9}\left ( \frac{t^2}{P^2} \right )+\frac{5}{81}\left ( \frac{t^3}{P^3} \right )+\dots\\ &=1+\frac{t}{3P}-\frac{t^2}{9P^2}+ \frac{5t^3}{81P^3}+\dots\\ \end{align*} {/eq}


Learn more about this topic:

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How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16
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