Simplify (x+iy)^3

${(x + i y)}^{3}$

Use the Binomial Theorem.

$x^{3} + 3 x^{2} (i y) + 3 x {(i y)}^{2} + {(i y)}^{3}$

Simplify each term.

Rewrite using the commutative property of multiplication.

$x^{3} + 3 \cdot i x^{2} y + 3 x {(i y)}^{2} + {(i y)}^{3}$

Apply the product rule to $i y$ .

$x^{3} + 3 i x^{2} y + 3 x (i^{2} y^{2}) + {(i y)}^{3}$

Rewrite using the commutative property of multiplication.

$x^{3} + 3 i x^{2} y + 3 \cdot i^{2} x y^{2} + {(i y)}^{3}$

Rewrite $i^{2}$ as $- 1$ .

$x^{3} + 3 i x^{2} y + 3 \cdot - 1 x y^{2} + {(i y)}^{3}$

Multiply $3$ by $- 1$ .

$x^{3} + 3 i x^{2} y - 3 x y^{2} + {(i y)}^{3}$

Apply the product rule to $i y$ .

$x^{3} + 3 i x^{2} y - 3 x y^{2} + i^{3} y^{3}$

Factor out $i^{2}$ .

$x^{3} + 3 i x^{2} y - 3 x y^{2} + i^{2} \cdot i y^{3}$

Rewrite $i^{2}$ as $- 1$ .

$x^{3} + 3 i x^{2} y - 3 x y^{2} - 1 \cdot i y^{3}$

Rewrite $- 1 i$ as $- i$ .

$x^{3} + 3 i x^{2} y - 3 x y^{2} - i y^{3}$