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To solve the equation [tex]\( -\left(-2 x^3\right)^6 = -12 x^9 \)[/tex], let's work through the solution step-by-step.
### Step 1: Simplify the Left-Hand Side
First, simplify the expression inside the parentheses:
[tex]\[ (-2 x^3)^6 \][/tex]
Since we are raising a product to a power, apply the power to both terms inside:
[tex]\[ (-2)^6 \times (x^3)^6 \][/tex]
Compute [tex]\((-2)^6\)[/tex]:
[tex]\[ (-2)^6 = 64 \][/tex]
Next, compute [tex]\((x^3)^6\)[/tex]:
[tex]\[ (x^3)^6 = x^{3 \times 6} = x^{18} \][/tex]
Now combine these results:
[tex]\[ (-2 x^3)^6 = 64 x^{18} \][/tex]
Remember, there is a negative sign outside the parentheses, so:
[tex]\[ -(-2 x^3)^6 = -64 x^{18} \][/tex]
### Step 2: Compare with the Right-Hand Side
The given right-hand side of the equation is:
[tex]\[ -12 x^9 \][/tex]
### Step 3: Set Up the Equation
Now, set the simplified left-hand side equal to the right-hand side:
[tex]\[ -64 x^{18} = -12 x^9 \][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
First, divide both sides by -1 to eliminate the negative signs:
[tex]\[ 64 x^{18} = 12 x^9 \][/tex]
Next, divide both sides by 12 to isolate the term with the highest power of [tex]\(x\)[/tex]:
[tex]\[ \frac{64}{12} x^{18} = x^9 \][/tex]
Simplify the coefficient:
[tex]\[ \frac{64}{12} = \frac{16}{3} \][/tex]
So, the equation becomes:
[tex]\[ \frac{16}{3} x^{18} = x^9 \][/tex]
To further isolate [tex]\(x\)[/tex], divide both sides by [tex]\( x^9 \)[/tex]:
[tex]\[ \frac{16}{3} x^9 = 1 \][/tex]
Now, solve for [tex]\(x^9\)[/tex]:
[tex]\[ x^9 = \frac{3}{16} \][/tex]
Take the ninth root of both sides to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \left(\frac{3}{16}\right)^{1/9} \][/tex]
### Step 5: Find the General Solutions
The above expression gives us one solution, but since we're dealing with polynomial equations of higher degrees, there could be additional complex solutions. For [tex]\(x\)[/tex], the general solutions can be expressed in terms of roots of unity.
Therefore, the complete set of solutions can be written as:
[tex]\[ x = 0, \quad \left(\frac{3}{16}\right)^{1/9}, \quad -\left(\frac{3}{16}\right)^{1/9}, \quad \text{and other complex solutions within the 9th roots of unity.} \][/tex]
This detailed set of solutions is provided by:
[tex]\[ x = [0, 2^{5/9} \cdot 3^{1/9}/2, -2^{5/9} \cdot 3^{1/9}/4 - 2^{5/9} \cdot 3^{11/18}i/4, -2^{5/9} \cdot 3^{1/9}/4 + 2^{5/9} \cdot 3^{11/18}i/4, -2^{5/9} \cdot 3^{1/9}\cos(\pi/9)/2 - 2^{5/9} \cdot 3^{1/9}i \sin(\pi/9)/2, -2^{5/9} \cdot 3^{1/9}\cos(\pi/9)/2 + 2^{5/9} \cdot 3^{1/9}i \sin(\pi/9)/2, 2^{5/9} \cdot 3^{1/9}\cos(2\pi/9)/2 - 2^{5/9} \cdot 3^{1/9}i \sin(2\pi/9)/2, 2^{5/9} \cdot 3^{1/9}\cos(2\pi/9)/2 + 2^{5/9} \cdot 3^{1/9}i \sin(2\pi/9)/2, 2^{5/9} \cdot 3^{1/9}\cos(4\pi/9)/2 - 2^{5/9} \cdot 3^{1/9}i \sin(4\pi/9)/2, 2^{5/9} \cdot 3^{1/9}\cos(4\pi/9)/2 + 2^{5/9} \cdot 3^{1/9}i \sin(4\pi/9)/2] \][/tex]
### Step 1: Simplify the Left-Hand Side
First, simplify the expression inside the parentheses:
[tex]\[ (-2 x^3)^6 \][/tex]
Since we are raising a product to a power, apply the power to both terms inside:
[tex]\[ (-2)^6 \times (x^3)^6 \][/tex]
Compute [tex]\((-2)^6\)[/tex]:
[tex]\[ (-2)^6 = 64 \][/tex]
Next, compute [tex]\((x^3)^6\)[/tex]:
[tex]\[ (x^3)^6 = x^{3 \times 6} = x^{18} \][/tex]
Now combine these results:
[tex]\[ (-2 x^3)^6 = 64 x^{18} \][/tex]
Remember, there is a negative sign outside the parentheses, so:
[tex]\[ -(-2 x^3)^6 = -64 x^{18} \][/tex]
### Step 2: Compare with the Right-Hand Side
The given right-hand side of the equation is:
[tex]\[ -12 x^9 \][/tex]
### Step 3: Set Up the Equation
Now, set the simplified left-hand side equal to the right-hand side:
[tex]\[ -64 x^{18} = -12 x^9 \][/tex]
### Step 4: Solve for [tex]\(x\)[/tex]
First, divide both sides by -1 to eliminate the negative signs:
[tex]\[ 64 x^{18} = 12 x^9 \][/tex]
Next, divide both sides by 12 to isolate the term with the highest power of [tex]\(x\)[/tex]:
[tex]\[ \frac{64}{12} x^{18} = x^9 \][/tex]
Simplify the coefficient:
[tex]\[ \frac{64}{12} = \frac{16}{3} \][/tex]
So, the equation becomes:
[tex]\[ \frac{16}{3} x^{18} = x^9 \][/tex]
To further isolate [tex]\(x\)[/tex], divide both sides by [tex]\( x^9 \)[/tex]:
[tex]\[ \frac{16}{3} x^9 = 1 \][/tex]
Now, solve for [tex]\(x^9\)[/tex]:
[tex]\[ x^9 = \frac{3}{16} \][/tex]
Take the ninth root of both sides to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \left(\frac{3}{16}\right)^{1/9} \][/tex]
### Step 5: Find the General Solutions
The above expression gives us one solution, but since we're dealing with polynomial equations of higher degrees, there could be additional complex solutions. For [tex]\(x\)[/tex], the general solutions can be expressed in terms of roots of unity.
Therefore, the complete set of solutions can be written as:
[tex]\[ x = 0, \quad \left(\frac{3}{16}\right)^{1/9}, \quad -\left(\frac{3}{16}\right)^{1/9}, \quad \text{and other complex solutions within the 9th roots of unity.} \][/tex]
This detailed set of solutions is provided by:
[tex]\[ x = [0, 2^{5/9} \cdot 3^{1/9}/2, -2^{5/9} \cdot 3^{1/9}/4 - 2^{5/9} \cdot 3^{11/18}i/4, -2^{5/9} \cdot 3^{1/9}/4 + 2^{5/9} \cdot 3^{11/18}i/4, -2^{5/9} \cdot 3^{1/9}\cos(\pi/9)/2 - 2^{5/9} \cdot 3^{1/9}i \sin(\pi/9)/2, -2^{5/9} \cdot 3^{1/9}\cos(\pi/9)/2 + 2^{5/9} \cdot 3^{1/9}i \sin(\pi/9)/2, 2^{5/9} \cdot 3^{1/9}\cos(2\pi/9)/2 - 2^{5/9} \cdot 3^{1/9}i \sin(2\pi/9)/2, 2^{5/9} \cdot 3^{1/9}\cos(2\pi/9)/2 + 2^{5/9} \cdot 3^{1/9}i \sin(2\pi/9)/2, 2^{5/9} \cdot 3^{1/9}\cos(4\pi/9)/2 - 2^{5/9} \cdot 3^{1/9}i \sin(4\pi/9)/2, 2^{5/9} \cdot 3^{1/9}\cos(4\pi/9)/2 + 2^{5/9} \cdot 3^{1/9}i \sin(4\pi/9)/2] \][/tex]
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