Partial Fractions: 1/(x³ − 1)
1/(x³ − 1) decomposes into (1/3)/(x - 1) + (-(1/3)x - 2/3)/(x^2 + x + 1). A worked irreducible quadratic example with the factorisation, the coefficient system, and an editable calculator.
1/(x³ − 1) = (1/3)/(x - 1) + (-(1/3)x - 2/3)/(x^2 + x + 1). x³ − 1 = (x − 1)(x² + x + 1); the quadratic has no real roots so it keeps a linear numerator.
e.g. 3x+11
Expanded or factored, e.g. (x-2)(x+3) or x^2+x-6
The decomposition
1/(x³ − 1) = (1/3)/(x - 1) + (-(1/3)x - 2/3)/(x^2 + x + 1)
Denominator factors: (x - 1) × (x^2 + x + 1)
This is a irreducible quadratic problem. x³ − 1 = (x − 1)(x² + x + 1); the quadratic has no real roots so it keeps a linear numerator.
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Frequently asked questions
How does 1/(x³ − 1) decompose? ▾
1/(x³ − 1) = (1/3)/(x - 1) + (-(1/3)x - 2/3)/(x^2 + x + 1). x³ − 1 = (x − 1)(x² + x + 1); the quadratic has no real roots so it keeps a linear numerator.
How was the denominator factored? ▾
It factors as (x - 1) × (x^2 + x + 1), which sets the shape of the decomposition.
How do I check the answer? ▾
Add the resulting fractions back over a common denominator — you should recover the original expression. Substituting a convenient value of x into both sides is a quicker spot check.
About this Calculator
1/(x³ − 1) decomposes into (1/3)/(x - 1) + (-(1/3)x - 2/3)/(x^2 + x + 1). A worked irreducible quadratic example with the factorisation, the coefficient system, and an editable calculator.