Applying the Perfect Cube Identity

The perfect cube forms $(x+y)^3$ and $( x-y)^3$ come up a lot in algebra. We will go over how to expand them in the examples below, but you should also take some time to store these forms in memory, since you'll see them often:

$\begin{aligned} (x+y)^3 &= x^3 + 3x^2 y + 3 x y^2 + y^3 \\ (x-y)^3 &= x^3 - 3x^2 y + 3 x y^2 - y^3. \end{aligned}$

We will also look at the sum and difference of cubes, and see various ways in which we can apply them:

$\begin{aligned} x^3 - y^3 &= (x-y) ( x^2 + xy + y^2) \\ x^3 + y^3 &= (x+y) ( x^2 - xy + y^2). \end{aligned}$

Expand $(x+1) ^3$.

---

Expanding out, we obtain

$(x+1)^3 = x^3 + 3 \times x^2 \times 1 + 3 \times x \times 1^2 + 1^3 = x^3 + 3x^2 + 3 x + 1 .\ _\square$

What is $(a-2b) ^ 3$?

---

Expanding out, we obtain

$\begin{aligned} ( a - 2b)^3 &= a^3 - 3 \times a^2 \times (2b) + 3 \times a \times (2b)^2 - (2b)^3 \\ &= a^3 - 6a^2b + 12a b^2 - 8 b^3.\ _\square \end{aligned}$

Factorize $x^3 + 8$.

---

We recognize that this is the sum of $x^3$ and $2^3$. Hence, by the sum of cubes factorization, we obtain

$x^3 + 8 = ( x + 2) (x ^2 - 2x + 2^2) = (x+2)( x^2 - 2x + 4 ).\ _\square$

What is $(x+y)^3 + (x-y)^3 ?$

---

Solution 1: If we look at the expansion of these terms, we see that the second and fourth terms will cancel out, and the first and third terms will be combined. Thus, we obtain

$( x+y)^3 + (x-y)^3 = 2 x^3 + 6 xy^2.\ _\square$

Solution 2: If we treat this as the sum of 2 cubes, we will obtain

$\begin{aligned} ( x+y)^3 + (x-y)^3 & = \big[ (x+y) + (x-y) \big] \big[ (x+y)^2 - (x+y)(x-y) + (x-y)^2 \big] \\ & = 2x \times \big[ x^2 + 3y^2 \big] \\ & = 2x^3 + 6xy^2 .\ _\square \end{aligned}$