Question 6
Given that $$ (8 + 3x)^{\frac{1}{3}} $$ can be expressed in the form $$ p(1 + qx)^{\frac{1}{3}} $$ where $p$ and $q$ are constants,
(a) find the value of $p$ and the value of $q$
(2)
(b) Hence, expand $$ (8 + 3x)^{\frac{1}{3}} $$ in ascending powers of $x$ up to and including the term in $x^2$, expressing each coefficient as an exact fraction in its lowest terms.
(3)
Using the expansion found in part (b) with a suitable value of $x$
(c) show that $$ \sqrt[3]{9} \approx \frac{599}{288} $$
(2)
Problem and Solution
We are given:
$$ (8 + 3x)^{\frac{1}{3}} = p(1 + qx)^{\frac{1}{3}} $$where $p$ and $q$ are constants.
Part (a): Finding $p$ and $q$
Factorize $8$ from $(8 + 3x)^{\frac{1}{3}}$:
$$ (8 + 3x)^{\frac{1}{3}} = 8^{\frac{1}{3}} \left(1 + \frac{3x}{8}\right)^{\frac{1}{3}} $$Since $8^{\frac{1}{3}} = 2$, we can write:
$$ (8 + 3x)^{\frac{1}{3}} = 2 \left(1 + \frac{3x}{8}\right)^{\frac{1}{3} } $$Comparing with $p(1 + qx)^{\frac{1}{3}}$, we find:
$$ p = 2, \quad q = \frac{3}{8}. $$Part (b): Expansion of $(8 + 3x)^{\frac{1}{3}$
Using the Binomial Theorem for fractional powers, expand $(1 + qx)^{\frac{1}{3}}$ up to the $x^2$ term:
$$ (1 + qx)^{\frac{1}{3}} = 1 + \frac{\frac{1}{3} \cdot qx}{1!} + \frac{\frac{1}{3} \left(\frac{1}{3} - 1\right) (qx)^2}{2!} + \dots $$Substitute $q = \frac{3}{8}$:
$$ (1 + qx)^{\frac{1}{3}} = 1 + \frac{\frac{1}{3} \cdot \frac{3}{8}x}{1} + \frac{\frac{1}{3} \left(\frac{1}{3} - 1\right) \left(\frac{3}{8}x\right)^2}{2} $$Simplify:
$$ \text{Linear term:} \quad \frac{\frac{1}{3} \cdot \frac{3}{8}x}{1} = \frac{1}{8}x $$ $$ \text{Quadratic term:} \quad \frac{\frac{1}{3} \left(\frac{1}{3} - 1\right) \left(\frac{3}{8}x\right)^2}{2} = \frac{\frac{1}{3} \cdot \left(-\frac{2}{3}\right) \cdot \frac{9}{64}x^2}{2} = -\frac{1}{64}x^2 $$Thus:
$$ (1 + qx)^{\frac{1}{3}} = 1 + \frac{1}{8}x - \frac{1}{64}x^2 $$Now multiply by $p = 2$:
$$ (8 + 3x)^{\frac{1}{3}} = 2 \left(1 + \frac{1}{8}x - \frac{1}{64}x^2\right) = 2 + \frac{1}{4}x - \frac{1}{32}x^2 $$Part (c): Approximation of $\sqrt[3]{9}$
To approximate $\sqrt[3]{9}$, use the expansion with $x = \frac{1}{3}$ (so that $8 + 3x = 9$):
$$ (8 + 3x)^{\frac{1}{3}} \approx 2 + \frac{1}{4}\left(\frac{1}{3}\right) - \frac{1}{32}\left(\frac{1}{3}\right)^2 $$Compute each term:
$$ \text{Constant term:} \quad 2 $$ $$ \text{Linear term:} \quad \frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12} $$ $$ \text{Quadratic term:} \quad \frac{1}{32} \cdot \frac{1}{9} = \frac{1}{288} $$Add these terms:
$$ \sqrt[3]{9} \approx 2 + \frac{1}{12} - \frac{1}{288} $$Convert all terms to have a denominator of $288$:
$$ 2 = \frac{576}{288}, \quad \frac{1}{12} = \frac{24}{288}, \quad \frac{1}{288} = \frac{1}{288} $$Add them:
$$ \sqrt[3]{9} \approx \frac{576 + 24 - 1}{288} = \frac{599}{288} $$Final Answers
- $p = 2$, $q = \frac{3}{8}$
- Expansion: $$ (8 + 3x)^{\frac{1}{3}} \approx 2 + \frac{1}{4}x - \frac{1}{32}x^2 $$
- Approximation: $$ \sqrt[3]{9} \approx \frac{599}{288} $$
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