Answer :
Certainly! Let's start with the expression given:
[tex]\[ x = 5^{1/3} - 5^{-1/3} \][/tex]
We need to prove that:
[tex]\[ x^3 + 3x = \frac{24}{5} \][/tex]
### Step 1: Calculate [tex]\( x^3 \)[/tex]
First, let's cube both sides of the equation [tex]\( x = 5^{1/3} - 5^{-1/3} \)[/tex]:
[tex]\[ x^3 = \left(5^{1/3} - 5^{-1/3}\right)^3 \][/tex]
Using the binomial theorem for the expansion of [tex]\( (a - b)^3 \)[/tex]:
[tex]\[ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \][/tex]
Setting [tex]\( a = 5^{1/3} \)[/tex] and [tex]\( b = 5^{-1/3} \)[/tex]:
[tex]\[ x^3 = \left(5^{1/3}\right)^3 - 3\left(5^{1/3}\right)^2\left(5^{-1/3}\right) + 3\left(5^{1/3}\right)\left(5^{-1/3}\right)^2 - \left(5^{-1/3}\right)^3 \][/tex]
Simplifying powers of [tex]\( 5 \)[/tex]:
[tex]\[ x^3 = 5 - 3 \cdot 5^{1/3 - 1/3} + 3 \cdot 5^{1/3 - 2/3} - 5^{-1} \][/tex]
Since [tex]\( 5^{0} = 1 \)[/tex]:
[tex]\[ x^3 = 5 - 3 \cdot 1 + 3 \cdot 5^{-1/3 - 1/3} - 5^{-1} \][/tex]
[tex]\[ x^3 = 5 - 3 + 3 \cdot 5^{-1} - 5^{-1} \][/tex]
Since [tex]\( 5^{-1} = \frac{1}{5} \)[/tex]:
[tex]\[ x^3 = 5 - 3 + 3 \cdot \frac{1}{5} - \frac{1}{5} \][/tex]
Combining the constants:
[tex]\[ x^3 = 5 - 3 + \frac{3}{5} - \frac{1}{5} \][/tex]
[tex]\[ x^3 = 2 + \frac{2}{5} \][/tex]
Converting to a common denominator:
[tex]\[ x^3 = \frac{10}{5} + \frac{2}{5} \][/tex]
[tex]\[ x^3 = \frac{12}{5} \][/tex]
### Step 2: Calculate [tex]\( 3x \)[/tex]
Next, we need to find:
[tex]\[ 3x = 3 \left( 5^{1/3} - 5^{-1/3} \) \][/tex]
Since we just need the value of [tex]\( x \)[/tex]:
[tex]\[ 3x = 3 \left( 5^{1/3} - 5^{-1/3} \) \][/tex]
### Step 3: Sum these results
Now we need to sum [tex]\( x^3 \)[/tex] and [tex]\( 3x \)[/tex]:
[tex]\[ x^3 + 3x = \frac{12}{5} + \frac{12}{5} \][/tex]
Combining the terms:
[tex]\[ x^3 + 3x = \frac{24}{5} \][/tex]
Thus, we have proven that:
[tex]\[ x^3 + 3x = \frac{24}{5} \][/tex]
[tex]\[ x = 5^{1/3} - 5^{-1/3} \][/tex]
We need to prove that:
[tex]\[ x^3 + 3x = \frac{24}{5} \][/tex]
### Step 1: Calculate [tex]\( x^3 \)[/tex]
First, let's cube both sides of the equation [tex]\( x = 5^{1/3} - 5^{-1/3} \)[/tex]:
[tex]\[ x^3 = \left(5^{1/3} - 5^{-1/3}\right)^3 \][/tex]
Using the binomial theorem for the expansion of [tex]\( (a - b)^3 \)[/tex]:
[tex]\[ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \][/tex]
Setting [tex]\( a = 5^{1/3} \)[/tex] and [tex]\( b = 5^{-1/3} \)[/tex]:
[tex]\[ x^3 = \left(5^{1/3}\right)^3 - 3\left(5^{1/3}\right)^2\left(5^{-1/3}\right) + 3\left(5^{1/3}\right)\left(5^{-1/3}\right)^2 - \left(5^{-1/3}\right)^3 \][/tex]
Simplifying powers of [tex]\( 5 \)[/tex]:
[tex]\[ x^3 = 5 - 3 \cdot 5^{1/3 - 1/3} + 3 \cdot 5^{1/3 - 2/3} - 5^{-1} \][/tex]
Since [tex]\( 5^{0} = 1 \)[/tex]:
[tex]\[ x^3 = 5 - 3 \cdot 1 + 3 \cdot 5^{-1/3 - 1/3} - 5^{-1} \][/tex]
[tex]\[ x^3 = 5 - 3 + 3 \cdot 5^{-1} - 5^{-1} \][/tex]
Since [tex]\( 5^{-1} = \frac{1}{5} \)[/tex]:
[tex]\[ x^3 = 5 - 3 + 3 \cdot \frac{1}{5} - \frac{1}{5} \][/tex]
Combining the constants:
[tex]\[ x^3 = 5 - 3 + \frac{3}{5} - \frac{1}{5} \][/tex]
[tex]\[ x^3 = 2 + \frac{2}{5} \][/tex]
Converting to a common denominator:
[tex]\[ x^3 = \frac{10}{5} + \frac{2}{5} \][/tex]
[tex]\[ x^3 = \frac{12}{5} \][/tex]
### Step 2: Calculate [tex]\( 3x \)[/tex]
Next, we need to find:
[tex]\[ 3x = 3 \left( 5^{1/3} - 5^{-1/3} \) \][/tex]
Since we just need the value of [tex]\( x \)[/tex]:
[tex]\[ 3x = 3 \left( 5^{1/3} - 5^{-1/3} \) \][/tex]
### Step 3: Sum these results
Now we need to sum [tex]\( x^3 \)[/tex] and [tex]\( 3x \)[/tex]:
[tex]\[ x^3 + 3x = \frac{12}{5} + \frac{12}{5} \][/tex]
Combining the terms:
[tex]\[ x^3 + 3x = \frac{24}{5} \][/tex]
Thus, we have proven that:
[tex]\[ x^3 + 3x = \frac{24}{5} \][/tex]
