We can factor both the sum and the difference of two cubes
Memorize this pattern
(A^3 + b^3) = A^2 - AB + b^2)
(A^3 - b^3) = A^2 + AB + b^2)
(x^3-1)/(x-1)
= (x-1)(x^2 + x + 1)/(x-1)
= x^2 + x + 1, x ≠ 1
How to simplify by factoring (x^3-1)/(x-1)?
I tried (x^2)(x-1)/(x-1), then (x+1)(x-1)(x+1)/(x-1).
2 answers
Apply difference of cubes formula:
x³ − y³ = ( x − y ) ( x² + x ∙ y + y² )
Replace y with 1
x³ − 1³ = ( x − 1 ) ( x² + x ∙ 1 + 1² )
x³ − 1 = ( x − 1 ) ( x² + x + 1 )
So:
( x³ − 1) / ( x - 1 ) = ( x − 1 ) ∙ ( x² + x + 1 ) / ( x - 1 ) = x² + x + 1
You also can use long division of polynomials.
Result is again:
( x³ − 1) / ( x - 1 ) = x² + x + 1
x³ − y³ = ( x − y ) ( x² + x ∙ y + y² )
Replace y with 1
x³ − 1³ = ( x − 1 ) ( x² + x ∙ 1 + 1² )
x³ − 1 = ( x − 1 ) ( x² + x + 1 )
So:
( x³ − 1) / ( x - 1 ) = ( x − 1 ) ∙ ( x² + x + 1 ) / ( x - 1 ) = x² + x + 1
You also can use long division of polynomials.
Result is again:
( x³ − 1) / ( x - 1 ) = x² + x + 1