# Thread: 8.24 is not like 1 + 3x

1. ## 8.24 is not like 1 + 3x

Hi guys,

now this is an easy one.

Given the expansion of $(1 + 3x)^{1/3} = 1 + x - x^2 + \frac{5}{3} x^3$

find $(8.24)^{1/3}$

so the problem is to get 8.24 into the form 1 + 3x i.e. x = 2.413 but it can't be that! Since the index of the expansion is 1/3 the binomial is only valid if mod x < 1/3.

Clearly I am missing something obvious!

2. Hi

$(8.24)^{\frac{1}{3}} = (8\times 1.03)^{\frac{1}{3}} = 2\cdot (1+3\cdot 0.01)^{\frac{1}{3}}$

3. Ah! That's it! Thankyou my friend - a very quick reply!

4. $8.24^{\frac{1}{3}}$
$=(8+.24)^{\frac{1}{3}}$
$=2(1+\frac{.24}{8})^{\frac{1}{3}}$
$=2(1+3\frac{.08}{8})^{\frac{1}{3}}$
$=2[1+3(.01)]^{\frac{1}{3}}$

5. Originally Posted by s_ingram
Hi guys,

now this is an easy one.

Given the expansion of $(1 + 3x)^{1/3} = 1 + x - x^2 + \frac{5}{3} x^3$

find $(8.24)^{1/3}$

so the problem is to get 8.24 into the form 1 + 3x i.e. x = 2.413 but it can't be that! Since the index of the expansion is 1/3 the binomial is only valid if mod x < 1/3.

Clearly I am missing something obvious!
From the given expression it follows that $(8 + 24x)^{1/3} = 2 (1 + 3x)^{1/3} = 2\left(1 + x - x^2 + \frac{5}{3} x^3\right)$.

Substitute $x = \frac{1}{100}$.