Use Pascal's Triangle to expand $$ \left(2 y^{2}-4\right)^{4} $$. Express your answer in simplest form.

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The expanded form is $$ 16y^8 - 128y^6 + 384y^4 - 512y^2 + 256 $$. To expand $$ (2y^2 - 4)^4 $$ using Pascal's Triangle, we first identify the coefficients from the 4th row of Pascal's Triangle, which are $$ 1, 4, 6, 4, 1 $$. The binomial expansion formula is given by: $$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k $$ In our case, we have $$ a = 2y^2 $$, $$ b = -4 $$, and $$ n = 4 $$. Thus, we can write: $$ (2y^2 - 4)^4 = \sum_{k=0}^{4} \binom{4}{k} (2y^2)^{4-k} (-4)^k $$ Now we will calculate each term in the expansion: 1. For $$ k = 0 $$: $$ \binom{4}{0} (2y^2)^{4} (-4)^{0} = 1 \cdot (2y^2)^{4} \cdot 1 = (2y^2)^{4} = 16y^8 $$ 2. For $$ k = 1 $$: $$ \binom{4}{1} (2y^2)^{3} (-4)^{1} = 4 \cdot (2y^2)^{3} \cdot (-4) = 4 \cdot 8y^6 \cdot (-4) = -128y^6 $$ 3. For $$ k = 2 $$: $$ \binom{4}{2} (2y^2)^{2} (-4)^{2} = 6 \cdot (2y^2)^{2} \cdot 16 = 6 \cdot 4y^4 \cdot 16 = 384y^4 $$ 4. For $$ k = 3 $$: $$ \binom{4}{3} (2y^2)^{1} (-4)^{3} = 4 \cdot (2y^2)^{1} \cdot (-64) = 4 \cdot 2y^2 \cdot (-64) = -512y^2 $$ 5. For $$ k = 4 $$: $$ \binom{4}{4} (2y^2)^{0} (-4)^{4} = 1 \cdot 1 \cdot 256 = 256 $$ Now, we combine all the terms: $$ (2y^2 - 4)^4 = 16y^8 - 128y^6 + 384y^4 - 512y^2 + 256 $$ Thus, the final expanded form of $$ (2y^2 - 4)^4 $$ is: $$ \boxed{16y^8 - 128y^6 + 384y^4 - 512y^2 + 256} $$

Use Pascal's Triangle to expand \( \left(2 y^{2}-4\right)^{4} \). Express your answer in simplest form.

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Use Pascal's Triangle to expand \( \left(2 y^{2}-4\right)^{4} \). Express your answer in simplest form.

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