act-dog

A. Correct. Rewrite the expression without exponents: ${\left(y+7\right)}^3=\left(y+7\right)\left(y+7\right)\left(y+7\right)$. Multiply the first $2$ binomials: $\left(y+7\right)\left(y+7\right)\left(y+7\right)=\left(y^2+14y+49\right)\left(y+7\right)$. Then, multiply the trinomial $\left(y^2+14y+49\right)$ by the binomial $\left(y+7\right)$ using the distributive property: $\left(y^2+14y+49\right)\left(y+7\right)=y\left(y^2+14y+49\right)+7\left(y^2+14y+49\right)$. Expand the expression and combine like terms: $y^3+14y^2+49y+7y^2+98y+343=y^3+21y^2+147y+343$. 

B. Incorrect. You may have correctly derived the expression $\left(y^2+14y+49\right)\left(y+7\right)$, but instead of multiplying the binomial by the trinomial, you may have multiplied $y$ by only $y^2$ and $7$ by only $49$ to get  $y\left(y^2\right)+14y+7\left(49\right)=y^3+14y+343$. Be sure to use the distributive property correctly and combine like terms.

C. Incorrect. You may have correctly derived the expression $\left(y^2+14y+49\right)\left(y+7\right)$, but instead of multiplying the binomial by the trinomial, you may have multiplied the first term by only $y$ to get $y\left(y^2\right)+14y+49=y^3+14y+49$. Be sure to use the distributive property and combine like terms.

D. Incorrect. You may have cubed the $y$ and the $7$ and then added the results to get $y^3+343$. Instead, rewrite the expression without exponents: ${\left(y+7\right)}^3=\left(y+7\right)\left(y+7\right)\left(y+7\right)$. Then, use the distributive property to expand the expression and combine like terms.