Question

If $x+y=4$ and $x^2+y^2=8$, find $x^3+y^3$.

Answer 1

16

Explanation:

We are given
`x+y=4`.

If we square both sides we obtain
`(x+y)^2=16`.

Recall
`(x+y)^2=x^2+2xy+y^2`.

So we have
`x^2+2xy+y^2=16`.

Reorder this so we can use the second given:
`x^2+y^2=8`.

`x^2+y^2+2xy=16`

`8+2xy=16`

Subtract 8 on both sides:

`2xy=8`

Divide both sides by 2:

`xy=4`

So we have the following so far:

`x+y=4`

`xy=4`

`x^2+y^2=8`

We are asked to find
`x^3+y^3`.

We will now cube both sides of the first given:

`x+y=4`

`(x+y)^3=64`

Expand left hand side:

`x^3+3x^2y+3xy^2+y^3=64`

Factor and reorder:

`x^3+y^3+3xy(x+y)=64`

Plug in values given and the value we found for
`xy`:

`x^3+y^3+3(4)(4)=64`

Simplify third term of left hand side:

`x^3+y^3+48=64`

Subtract 48 on both sides:

`x^3+y^3=16`