In part B we must perform the following operation:
[tex](5a^3+4a^2-3a+2)+(a^3-3a^2+3a-9)[/tex]The key here is to group the terms according to the power of a they have:
[tex](5a^3+4a^2-3a+2)+(a^3-3a^2+3a-9)=(5a^3+a^3)+(4a^2-3a^2)+(-3a+3a)+(2-9)[/tex]Then, we can use the distributive property of the multiplication but in reverse:
[tex]b\cdot a+c\cdot a=(b+c)\cdot a[/tex]If we do this in each of the terms between parenthesis we get:
[tex]\begin{gathered} (5a^3+a^3)+(4a^2-3a^2)+(-3a+3a)+(2-9)= \\ =(5+1)a^3+(4-3)a^2+(-3+3)a-7 \\ (5+1)a^3+(4-3)a^2+(-3+3)a-7=6a^3+a^2-7 \end{gathered}[/tex]Then the answer for part B is:
[tex]6a^3+a^2-7[/tex]In part C we must simplify:
[tex](4y^3-2y+9)-(2y^3-3y^2+4y+7)[/tex]Here is important to remember that a negative sign before a parenthesis means that you have to change the sign of all the terms inside it. Then we have:
[tex](4y^3-2y+9)-(2y^3-3y^2+4y+7)=4y^3-2y+9-2y^3+3y^2-4y-7[/tex]Now we can do the same thing we did in part B, we group the terms according to the powers of y:
[tex]4y^3-2y+9-2y^3+3y^2-4y-7=(4y^3-2y^3)+3y^2+(-2y-4y)+(9-7)[/tex]Then we apply the distributive property in reverse:
[tex]\begin{gathered} (4y^3-2y^3)+3y^2+(-2y-4y)+(9-7)=(4-2)y^3+3y^2+(-2-4)y+2 \\ (4-2)y^3+3y^2+(-2-4)y+2=2y^3+3y^2-6y+2 \end{gathered}[/tex]Then the answer for part C is:
[tex]2y^3+3y^2-6y+2[/tex]