Question

Factorise the following:

x³ + 3x²y + 3xy² + 9y³

Answer 1

Given: x³ + 3x²y + 3xy² + 9y³

Step-wise calculation:

1. Recognize the pattern: The expression looks like a cubic expansion.

Recall that (a + b)³ = a³ + 3a²b + 3ab² + b³

2. Compare the given expression with the expansion:

x³ + 3x²y + 3xy² + 9y³

Notice that the last term is (9y³), which can be rewritten as (3y)³.

3. Thus, the expression is like (x + 3y)³ but the last term in the expansion of (x + 3y)³ would be (3y)³ = 27y³, which is different.

4. Alternatively, try rewriting the expression grouping within the cubic expansion form.

Let’s look if the expression can be factored as a perfect cube of the form (x + ay)³:

(x + ay)³ = x³ + 3ax²y + 3a²xy² + a³y³

Matching the coefficients:

• 3a = 3 ⇒ a = 1
• 3a² = 3 ⇒ 3(1)² = 3 matches
• a³ = 9 ⇒ 1³ = 1 ≠ 9 does not match

So a = 1 does not satisfy the last term.

Try `a = root(3)(9)`, but that is complicated.

Alternatively, rewrite the expression grouping the terms:

x³ + 3x²y + 3xy² + 9y³ = (x³ + 3x²y + 3xy² + y³) + 8y³

Because we subtracted and added (y³) to make a perfect cube (x + y)³ = x³ + 3x²y + 3xy² + y³.

So, (x + y)³ + 8y³ = (x + y)³ + (2y)³.

Now it is sum of cubes:

a³ + b³ = (a + b)(a² – ab + b²)

where (a = x + y), (b = 2y).

Applying formula:

(x + y)³ + (2y)³ = ((x + y) + 2y) ((x + y)² – (x + y)(2y) + (2y)²)

Simplify the first factor:

(x + y) + 2y = x + 3y

Simplify the second factor:

(x + y)² – 2y(x + y) + 4y²

Calculate each term:

• (x + y)² = x² + 2xy + y²
• 2y(x + y) = 2xy + 2y²

So:

x² + 2xy + y² – (2xy + 2y²) + 4y² 

= x² + 2xy + y² – 2xy – 2y² + 4y²

= x² + (2xy – 2xy) + (y² – 2y² + 4y²) 

= x² + 0 + 3y²

= x² + 3y²

x³ + 3x²y + 3xy² + 9y³ = (x + 3y)(x² + 3y²)