Answer:
16
Step-by-step explanation:
We are given [tex]x+y=4[/tex].
If we square both sides we obtain [tex](x+y)^2=16[/tex].
Recall [tex](x+y)^2=x^2+2xy+y^2[/tex].
So we have [tex]x^2+2xy+y^2=16[/tex].
Reorder this so we can use the second given: [tex]x^2+y^2=8[/tex].
[tex]x^2+y^2+2xy=16[/tex]
[tex]8+2xy=16[/tex]
Subtract 8 on both sides:
[tex]2xy=8[/tex]
Divide both sides by 2:
[tex]xy=4[/tex]
So we have the following so far:
[tex]x+y=4[/tex]
[tex]xy=4[/tex]
[tex]x^2+y^2=8[/tex]
We are asked to find [tex]x^3+y^3[/tex].
We will now cube both sides of the first given:
[tex]x+y=4[/tex]
[tex](x+y)^3=64[/tex]
Expand left hand side:
[tex]x^3+3x^2y+3xy^2+y^3=64[/tex]
Factor and reorder:
[tex]x^3+y^3+3xy(x+y)=64[/tex]
Plug in values given and the value we found for [tex]xy[/tex]:
[tex]x^3+y^3+3(4)(4)=64[/tex]
Simplify third term of left hand side:
[tex]x^3+y^3+48=64[/tex]
Subtract 48 on both sides:
[tex]x^3+y^3=16[/tex]
