Math Problem Statement

Find the binomial expansion of (27 + 6x)^(1/3) up to and including the term in x^3, simplifying the coefficients. Then, given that cube root of 3 = 3/cube root of 9, use the binomial expansion to estimate the cube root of 3, giving the answer to five significant places.

Solution

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Part (i): Find the binomial expansion of (27+6x)13(27 + 6x)^{\frac{1}{3}} up to and including the term in x3x^3, simplifying the coefficients.

Part (ii): Given that 33=393\sqrt[3]{3} = \frac{3}{\sqrt[3]{9}}, use your binomial expansion in part (i) to obtain the value of 33\sqrt[3]{3}, giving the answer to five significant places.


Part (i): Binomial Expansion

The expression given is (27+6x)13(27 + 6x)^{\frac{1}{3}}.

To simplify the binomial expansion, we can factor out the 27:

(27+6x)13=2713(1+6x27)13=3(1+2x9)13(27 + 6x)^{\frac{1}{3}} = 27^{\frac{1}{3}} \left( 1 + \frac{6x}{27} \right)^{\frac{1}{3}} = 3 \left( 1 + \frac{2x}{9} \right)^{\frac{1}{3}}

Now, we use the binomial expansion formula for (1+u)n(1 + u)^n where u=2x9u = \frac{2x}{9} and n=13n = \frac{1}{3}:

(1+u)n=1+nu+n(n1)2!u2+n(n1)(n2)3!u3+(1 + u)^n = 1 + n u + \frac{n(n-1)}{2!} u^2 + \frac{n(n-1)(n-2)}{3!} u^3 + \dots

Expanding (1+2x9)13\left( 1 + \frac{2x}{9} \right)^{\frac{1}{3}}:

(1+2x9)13=1+132x9+13(131)2!(2x9)2+\left( 1 + \frac{2x}{9} \right)^{\frac{1}{3}} = 1 + \frac{1}{3} \cdot \frac{2x}{9} + \frac{\frac{1}{3} \cdot \left( \frac{1}{3} - 1 \right)}{2!} \cdot \left( \frac{2x}{9} \right)^2 + \dots

This simplifies to:

=1+2x272x2243+= 1 + \frac{2x}{27} - \frac{2x^2}{243} + \dots

Thus, multiplying by 3 (from the earlier factoring out of 27):

3(1+2x272x2243+)=3+2x92x281+3 \left( 1 + \frac{2x}{27} - \frac{2x^2}{243} + \dots \right) = 3 + \frac{2x}{9} - \frac{2x^2}{81} + \dots

Including up to the x3x^3 term, we have:

3+2x92x281+3 + \frac{2x}{9} - \frac{2x^2}{81} + \dots


Part (ii): Using Binomial Expansion to Estimate 33\sqrt[3]{3}

We are given 33=393\sqrt[3]{3} = \frac{3}{\sqrt[3]{9}}.

We know that 93=(32)1/3=32/3\sqrt[3]{9} = (3^2)^{1/3} = 3^{2/3}.

To estimate 33\sqrt[3]{3}, we can use the binomial expansion we found in part (i). By setting x=1x = 1, we substitute this into the expansion from part (i):

933+21921281=3+29281=3+0.22220.0247=3.1975\sqrt[3]{9} \approx 3 + \frac{2 \cdot 1}{9} - \frac{2 \cdot 1^2}{81} = 3 + \frac{2}{9} - \frac{2}{81} = 3 + 0.2222 - 0.0247 = 3.1975

Therefore,

[ \sqrt[3]{3} = \frac{3}{\sqrt[3]{9}} \approx \frac{3}{3.1975} = 0.9383 ]

Finally, rounding to five significant places:

330.93828\sqrt[3]{3} \approx 0.93828


Would you like more details on any step?

Here are some related questions:

  1. How do we derive the general binomial expansion for fractional powers?
  2. Why is binomial expansion useful for approximations like these?
  3. What happens if we include more terms in the binomial expansion?
  4. How can we refine the approximation further using calculus methods?
  5. Can we use this method to approximate other roots, such as 34\sqrt[4]{3}?

Tip: When using binomial expansions, remember to watch for how many terms you need to include for the desired accuracy.

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Math Problem Analysis

Mathematical Concepts

Binomial Expansion
Algebra
Approximations

Formulas

Binomial expansion formula: (1 + u)^n = 1 + n*u + (n(n-1)/2!) * u^2 + (n(n-1)(n-2)/3!) * u^3 + ...
Cube roots approximation

Theorems

Binomial Theorem

Suitable Grade Level

Undergraduate Mathematics or Advanced High School (Grades 11-12)