Question

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

Explanation:

Step1: Recall Pascal's Triangle for power 3

The coefficients for $(a + b)^3$ from Pascal's Triangle are 1, 3, 3, 1.

Step2: Identify a and b

Here $a = 3z$ and $b=3y^{2}$.

Step3: Apply the binomial - expansion formula $(a + b)^3=a^{3}+3a^{2}b + 3ab^{2}+b^{3}$

Substitute $a = 3z$ and $b = 3y^{2}$ into the formula:

• For the first term $a^{3}$: $(3z)^{3}=27z^{3}$.
• For the second term $3a^{2}b$: $3\times(3z)^{2}\times(3y^{2})=3\times9z^{2}\times3y^{2}=81z^{2}y^{2}$.
• For the third term $3ab^{2}$: $3\times(3z)\times(3y^{2})^{2}=3\times3z\times9y^{4}=81zy^{4}$.
• For the fourth term $b^{3}$: $(3y^{2})^{3}=27y^{6}$.

Answer:

$27z^{3}+81z^{2}y^{2}+81zy^{4}+27y^{6}$