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Let's simplify and expand the expression [tex]\(\left(2u + \frac{3}{u} - 1\right)^2\)[/tex] step-by-step.
### Step 1: Understand the Expression Inside the Square
We have the expression inside the square as [tex]\(2u + \frac{3}{u} - 1\)[/tex].
### Step 2: Apply the Square Formula
Recall the algebraic identity for squaring a trinomial:
[tex]\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc \][/tex]
In our case, [tex]\(a = 2u\)[/tex], [tex]\(b = \frac{3}{u}\)[/tex], and [tex]\(c = -1\)[/tex]. So we can write:
[tex]\[ (2u + \frac{3}{u} - 1)^2 \][/tex]
### Step 3: Compute Individual Squares and Cross-terms
1. Calculate [tex]\(a^2\)[/tex]:
[tex]\[ (2u)^2 = 4u^2 \][/tex]
2. Calculate [tex]\(b^2\)[/tex]:
[tex]\[ \left(\frac{3}{u}\right)^2 = \frac{9}{u^2} \][/tex]
3. Calculate [tex]\(c^2\)[/tex]:
[tex]\[ (-1)^2 = 1 \][/tex]
4. Calculate [tex]\(2ab\)[/tex]:
[tex]\[ 2(2u) \left(\frac{3}{u}\right) = 2 \cdot 2u \cdot \frac{3}{u} = 2 \cdot 6 = 6 \][/tex]
5. Calculate [tex]\(2ac\)[/tex]:
[tex]\[ 2(2u) \cdot (-1) = 2 \cdot 2u \cdot -1 = -4u \][/tex]
6. Calculate [tex]\(2bc\)[/tex]:
[tex]\[ 2 \left( \frac{3}{u} \right) \cdot (-1) = 2 \cdot \frac{3}{u} \cdot -1 = -\frac{6}{u} \][/tex]
### Step 4: Sum up All the Parts
Combining all these results together, we get the expanded expression:
[tex]\[ 4u^2 + \frac{9}{u^2} + 1 + 6 - 4u - \frac{6}{u} \][/tex]
### Step 5: Simplify the Expression
Collect and combine like terms:
[tex]\[ 4u^2 - 4u + 1 + 6 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
[tex]\[ 4u^2 - 4u + 7 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
Thus, the expanded form of [tex]\(\left(2u + \frac{3}{u} - 1\right)^2\)[/tex] is:
[tex]\[ 4u^2 - 4u + 7 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
### Conclusion
We can summarize our results as the original and expanded forms:
[tex]\[ \left(2u + \frac{3}{u} - 1\right)^2 = \left(2u + \frac{3}{u} - 1\right)^2 \][/tex]
[tex]\[ 4u^2 - 4u + 7 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
So, the original expression [tex]\(\left(2u + \frac{3}{u} - 1\right)^2\)[/tex] simplifies and expands to:
[tex]\[ (2u - 1 + \frac{3}{u})^2 = 4u^2 - 4u + 7 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
### Step 1: Understand the Expression Inside the Square
We have the expression inside the square as [tex]\(2u + \frac{3}{u} - 1\)[/tex].
### Step 2: Apply the Square Formula
Recall the algebraic identity for squaring a trinomial:
[tex]\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc \][/tex]
In our case, [tex]\(a = 2u\)[/tex], [tex]\(b = \frac{3}{u}\)[/tex], and [tex]\(c = -1\)[/tex]. So we can write:
[tex]\[ (2u + \frac{3}{u} - 1)^2 \][/tex]
### Step 3: Compute Individual Squares and Cross-terms
1. Calculate [tex]\(a^2\)[/tex]:
[tex]\[ (2u)^2 = 4u^2 \][/tex]
2. Calculate [tex]\(b^2\)[/tex]:
[tex]\[ \left(\frac{3}{u}\right)^2 = \frac{9}{u^2} \][/tex]
3. Calculate [tex]\(c^2\)[/tex]:
[tex]\[ (-1)^2 = 1 \][/tex]
4. Calculate [tex]\(2ab\)[/tex]:
[tex]\[ 2(2u) \left(\frac{3}{u}\right) = 2 \cdot 2u \cdot \frac{3}{u} = 2 \cdot 6 = 6 \][/tex]
5. Calculate [tex]\(2ac\)[/tex]:
[tex]\[ 2(2u) \cdot (-1) = 2 \cdot 2u \cdot -1 = -4u \][/tex]
6. Calculate [tex]\(2bc\)[/tex]:
[tex]\[ 2 \left( \frac{3}{u} \right) \cdot (-1) = 2 \cdot \frac{3}{u} \cdot -1 = -\frac{6}{u} \][/tex]
### Step 4: Sum up All the Parts
Combining all these results together, we get the expanded expression:
[tex]\[ 4u^2 + \frac{9}{u^2} + 1 + 6 - 4u - \frac{6}{u} \][/tex]
### Step 5: Simplify the Expression
Collect and combine like terms:
[tex]\[ 4u^2 - 4u + 1 + 6 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
[tex]\[ 4u^2 - 4u + 7 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
Thus, the expanded form of [tex]\(\left(2u + \frac{3}{u} - 1\right)^2\)[/tex] is:
[tex]\[ 4u^2 - 4u + 7 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
### Conclusion
We can summarize our results as the original and expanded forms:
[tex]\[ \left(2u + \frac{3}{u} - 1\right)^2 = \left(2u + \frac{3}{u} - 1\right)^2 \][/tex]
[tex]\[ 4u^2 - 4u + 7 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
So, the original expression [tex]\(\left(2u + \frac{3}{u} - 1\right)^2\)[/tex] simplifies and expands to:
[tex]\[ (2u - 1 + \frac{3}{u})^2 = 4u^2 - 4u + 7 - \frac{6}{u} + \frac{9}{u^2} \][/tex]
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