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

User Mgarg
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1 Answer

7 votes

Answer:

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

User Omar N Shamali
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6.5k points