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 $b = 3y^{2}$:
$(3z)^{3}+3\times(3z)^{2}\times(3y^{2})+3\times(3z)\times(3y^{2})^{2}+(3y^{2})^{3}$

Step4: Simplify each term

$(3z)^{3}=3^{3}z^{3}=27z^{3}$;
$3\times(3z)^{2}\times(3y^{2})=3\times9z^{2}\times3y^{2}=81z^{2}y^{2}$;
$3\times(3z)\times(3y^{2})^{2}=3\times3z\times9y^{4}=81zy^{4}$;
$(3y^{2})^{3}=3^{3}y^{6}=27y^{6}$.

Step5: Combine the terms

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

Answer:

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