Given x + y = 3 and xy = 3, Find x³ + y³

Assuming the notation means $x+y=3$ and $xy=3$, you can find $x^3+y^3$ using an identity that relates the cubes to the sum and product. The calculation gives $0$. There is one important qualification: no real numbers meet both conditions, although complex numbers do.

Use the sum-of-cubes identity

Expand $(x+y)^3$:

$$(x+y)^3=x^3+3x^2y+3xy^2+y^3.$$

The middle terms combine as $3x^2y+3xy^2=3xy(x+y)$. Rearranging gives

$$x^3+y^3=(x+y)^3-3xy(x+y).$$

This identity is useful whenever you know the sum $x+y$ and product $xy$. It avoids finding $x$ and $y$ separately.

Substitute the given values

Use $x+y=3$ and $xy=3$:

$$\begin{aligned} x^3+y^3&=(x+y)^3-3xy(x+y)\\ &=3^3-3(3)(3)\\ &=27-27\\ &=0. \end{aligned}$$

So the algebraic value is 0.

Check whether real values are possible

The calculation above follows from the stated sum and product, but it does not establish that real numbers $x$ and $y$ with those values exist. For real numbers, $(x-y)^2$ cannot be negative. Since

$$(x-y)^2=(x+y)^2-4xy,$$

these conditions would require

$$(x-y)^2=3^2-4(3)=9-12=-3,$$

which is impossible for real $x$ and $y$. Equivalently, real numbers with sum 3 must have product at most $9/4$, not 3.

Thus, if the problem requires real $x$ and $y$, the conditions are inconsistent. If complex values are allowed, they are possible, and the value $x^3+y^3=0$ is attained.

Verify using the complex solutions

Treat $x$ and $y$ as the roots of a quadratic whose sum is 3 and product is 3. By Vieta’s formulas, that quadratic is

$$t^2-3t+3=0.$$

Its roots are

$$x,y=\frac{3\pm i\sqrt{3}}{2}.$$

These roots have sum 3 and product 3. Their cubes are $3\sqrt{3}i$ and $-3\sqrt{3}i$, respectively, so they sum to 0. This confirms the identity-based calculation with actual values.

Common mistakes

• Using the wrong identity: $x^3+y^3$ is not generally equal to $(x+y)^3$. The expansion has extra terms, which is why the correction $-3xy(x+y)$ is needed.
• Dropping the product: Knowing only $x+y=3$ is not enough to determine $x^3+y^3$. The value also depends on $xy$.
• Ignoring the real-number constraint: The algebraic substitution is valid, but it does not make inconsistent real conditions possible. Check whether the given sum and product can occur in the number system being used.

If only $x+y=3$ was intended and no product was specified, there is not a unique answer. For example, $(x,y)=(0,3)$ gives $x^3+y^3=27$, while $(x,y)=(1,2)$ gives $9$.