To expand the given expression, let's proceed step by step. The expression is:

$\left(\frac{v-u}{t} \cdot t \cdot v \right) - \left(\frac{v-u}{t} \cdot t \cdot u \right) + 2uv - 2u^2$

Step 1: Simplify the fractions involving $(v-u)/t$

• In $\frac{v-u}{t} \cdot t$, the $t$ cancels out, leaving just $(v-u)$.

So the expression becomes: $(v-u)v - (v-u)u + 2uv - 2u^2$

Step 2: Expand the terms

• Expanding $(v-u)v$: $(v-u)v = v^2 - uv$
• Expanding $(v-u)u$: $(v-u)u = uv - u^2$

Substitute these back into the expression: $(v^2 - uv) - (uv - u^2) + 2uv - 2u^2$

Step 3: Simplify

Distribute the negative sign for $-(uv - u^2)$: $v^2 - uv - uv + u^2 + 2uv - 2u^2$

Combine like terms:

• $v^2$ remains as is.
• Combine all $uv$ terms: $-uv - uv + 2uv = 0$.
• Combine $u^2$ terms: $u^2 - 2u^2 = -u^2$.

Thus, the simplified expression is: $v^2 - u^2$

Final Answer:

$v^2 - u^2$

Would you like further explanations or have any questions about the steps?

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Here are 5 related questions to explore:

1. What is the geometric interpretation of $v^2 - u^2$ as a difference of squares?
2. How does this expression relate to the expansion of $(v - u)(v + u)$?
3. What happens if $t = 0$ in the original equation? Why does it create a problem?
4. Can you factorize the expression $v^2 - u^2$?
5. How does this simplification apply in physics or other real-world contexts?

Tip: Remember, the difference of squares $a^2 - b^2$ always factors into ((a - b)(a + b)\