3. Use the distributive property to show that the equations are equal. Examples: \( a(b+c)=a \times b+a \times c \) \[ \begin{array}{l} a\left(b^{2}+c^{2}\right)=a \times b^{2}+a \times c^{2} \\ a(b-c)=a \times b-a \times c \\ a\left(b^{2}-c^{2}\right)=a \times b^{2}-a \times c^{2} \end{array} \] a. \( \left(b^{2}+c^{3}\right) d= \) b. \( c\left(b+d^{2}\right)= \) c. \( \left(b^{2}+d^{2}\right) \times c= \) E \( \square \) d. \( d \times\left(c+b^{2}\right)= \) e. \( \left(d^{2} \times b^{3}\right)+\left(d^{2} \times c^{3}\right)= \)

To show that the equations are equal using the distributive property, we will apply the property step by step for each expression provided. The distributive property states that \( a(b+c) = ab + ac \). Let's go through each part: ### a. \( (b^{2}+c^{3}) d \) Using the distributive property: \[ (b^{2}+c^{3}) d = b^{2}d + c^{3}d \] ### b. \( c(b+d^{2}) \) Using the distributive property: \[ c(b+d^{2}) = cb + cd^{2} \] ### c. \( (b^{2}+d^{2}) \times c \) Using the distributive property: \[ (b^{2}+d^{2}) \times c = b^{2}c + d^{2}c \] ### d. \( d \times (c+b^{2}) \) Using the distributive property: \[ d \times (c+b^{2}) = dc + db^{2} \] ### e. \( (d^{2} \times b^{3}) + (d^{2} \times c^{3}) \) This expression can be factored using the distributive property: \[ d^{2} \times b^{3} + d^{2} \times c^{3} = d^{2}(b^{3} + c^{3}) \] ### Summary of Results 1. \( (b^{2}+c^{3}) d = b^{2}d + c^{3}d \) 2. \( c(b+d^{2}) = cb + cd^{2} \) 3. \( (b^{2}+d^{2}) \times c = b^{2}c + d^{2}c \) 4. \( d \times (c+b^{2}) = dc + db^{2} \) 5. \( (d^{2} \times b^{3}) + (d^{2} \times c^{3}) = d^{2}(b^{3} + c^{3}) \) Each expression has been simplified using the distributive property, demonstrating that they are equal to their respective expanded forms.