Question

Factorise the following:

(x + y)³ – x – у

Answer 1

Given Expression: (x + y)³ – x – у

Step 1: Expand

(x + y)³(x + y)³ = x³ + 3x²y + 3xy² + y³

Step 2: Substitute the expansion back into the expression

(x + y)³ – x – y = x³ + 3x²y + 3xy² + y³ – x – y

Step 3: Rearrange terms (group terms)

x³ + y³ + 3x²y + 3xy² – x – y

Step 4: Recognize that

(3x²y + 3xy² = 3xy(x + y))

So the expression becomes x³ + y³ + 3xy(x + y) – x – y.

Step 5: Group terms as:

(x³ + y³) + 3xy(x + y) – (x + y)

Step 6: Factor (x + y) out of the last two terms: 

(x³ + y³) + (3xy – 1)(x + y)

Step 7: Recall the identity for sum of cubes: 

x³ + y³ = (x + y)(x² – xy + y²)

Hence: (x + y)(x² – xy + y²) + (3xy – 1)(x + y) = (x + y)[x² – xy + y² + 3xy – 1].

Step 8: Simplify inside the bracket:

x² – xy + y² + 3xy – 1 = x² + 2xy + y² – 1

x² – xy + y² + 3xy – 1 = (x + y)² – 1

Step 9: Recognize difference of squares:

(x + y)² – 1 = (x + y – 1)(x + y + 1)

Step 10: So the factorized form is:

(x + y)(x + y – 1)(x + y + 1)