Question

use pascals triangle to expand ((y^2 + z^2)^3). express your answer in simplest form.

Explanation:

Step1: Recall Pascal's Triangle for \(n = 3\)

Pascal's Triangle for the power \(n = 3\) gives the coefficients \(1, 3, 3, 1\). The binomial expansion formula is \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), and for \(n = 3\), the expansion is \(a^{3}+3a^{2}b + 3ab^{2}+b^{3}\).

Step2: Identify \(a\) and \(b\)

In the expression \((y^{2}+z^{2})^{3}\), let \(a = y^{2}\) and \(b = z^{2}\).

Step3: Substitute \(a\) and \(b\) into the expansion

Substitute \(a = y^{2}\) and \(b = z^{2}\) into \(a^{3}+3a^{2}b + 3ab^{2}+b^{3}\):

• For the first term: \(a^{3}=(y^{2})^{3}=y^{6}\) (using the power - of - a - power rule \((x^{m})^{n}=x^{mn}\))
• For the second term: \(3a^{2}b = 3(y^{2})^{2}(z^{2})=3y^{4}z^{2}\) (using the power - of - a - power rule and the product rule of exponents \(x^{m}\cdot x^{n}=x^{m + n}\))
• For the third term: \(3ab^{2}=3(y^{2})(z^{2})^{2}=3y^{2}z^{4}\) (using the power - of - a - power rule and the product rule of exponents)
• For the fourth term: \(b^{3}=(z^{2})^{3}=z^{6}\) (using the power - of - a - power rule)

Answer:

\(y^{6}+3y^{4}z^{2}+3y^{2}z^{4}+z^{6}\)