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Use Pascal's Triangle to expand
(2y^(2)+z^(2))^(3). Express your answer in simplest form.
Answer:

Use Pascal's Triangle to expand $\left(2 y^{2}+z^{2}\right)^{3}$ . Express your answer in simplest form. $\newline$ Answer:

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Pascal's triangle and the Binomial Theorem

Full solution

Q. Use Pascal's Triangle to expand

\left(2 y^{2}+z^{2}\right)^{3}

. Express your answer in simplest form.

\newline

Answer:

Identify Row: Identify the row of Pascal's Triangle that corresponds to the exponent of the binomial expansion. $\newline$ Since we are expanding $(2y^{2}+z^{2})^{3}$ , we need to look at the fourth row of Pascal's Triangle, which corresponds to the coefficients for a cubic expansion. $\newline$ The fourth row of Pascal's Triangle is $1, 3, 3, 1$ .
Write Terms: Write out each term of the expansion using the coefficients from Pascal's Triangle. $\newline$ The expansion will have four terms, and the coefficients will be $1$ , $3$ , $3$ , and $1$ , respectively. $\newline$ The terms will be: $\newline$ $1*(2y^{2})^{3}*(z^{2})^{0}$ , $\newline$ $3*(2y^{2})^{2}*(z^{2})^{1}$ , $\newline$ $3*(2y^{2})^{1}*(z^{2})^{2}$ , $\newline$ $1*(2y^{2})^{0}*(z^{2})^{3}$ .
Calculate Each Term: Calculate each term of the expansion. $\newline$ Now we will calculate each term: $\newline$ $1\times(2y^{2})^{3}\times(z^{2})^{0} = 1\times(8y^{6})\times(1) = 8y^{6}$ , $\newline$ $3\times(2y^{2})^{2}\times(z^{2})^{1} = 3\times(4y^{4})\times(z^{2}) = 12y^{4}z^{2}$ , $\newline$ $3\times(2y^{2})^{1}\times(z^{2})^{2} = 3\times(2y^{2})\times(z^{4}) = 6y^{2}z^{4}$ , $\newline$ $1\times(2y^{2})^{0}\times(z^{2})^{3} = 1\times(1)\times(z^{6}) = z^{6}$ .
Combine Final Form: Combine all the terms to write the final expanded form. $\newline$ The final expanded form of $(2y^{2}+z^{2})^{3}$ is: $\newline$ $8y^{6} + 12y^{4}z^{2} + 6y^{2}z^{4} + z^{6}$ .

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