How do I factor x^3 - 8?

x^3-8 = (x)^3 - (2)^3
a^3-b^3 = (a-b)(a^2+ab+b^2)
x^3-8 = (x-2)(x^2+2x+4)

a^3 - b^3 ≡ (a - b)(a^2 + ab + b^2)

x^3 - 8
= x^3 - 2^3
= (x - 2)(x^2 + 2x + 4)

x³ - 8 =

Notice that both terms are perfect cubes.
x³ = x * x * x = (x)³
8 = 2 * 2 * 2 = (2)³

Remember that you can factor the difference of two cubes.
a³ - b³ = (a - b)(a² + ab + b²)

Given: x³ - 8 = (x)³ - (2)³
Means: a = x, b = 2

Apply the factoring formula.
(x - 2)(x² + 2x + 4)

ANSWER: (x - 2)(x² + 2x + 4)

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Sometimes you'll come across polynomials that match a particular pattern.

HINT: Memorize these commonly occurring factoring formulas

Perfect square binomials:
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²

Difference of two perfect squares:
(a - b)(a + b) = a² - b²

Sum of two perfect cubes:
a³ + b³ = (a + b)(a² - ab + b²)

Difference of two perfect cubes:
a³ - b³ = (a - b)(a² + ab + b²)

Notice that the formulas for the perfect cubes are almost the same, but the signs are different. You can use the SOAP mnemonic to remember the signs.

S = The first sign is always the SAME as the sign between the 2 cubes
O = The second sign is always the OPPOSITE as the sign between the 2 cubes
AP = The last sign is ALWAYS POSITIVE

factor(x^3-8) = (x-2)*(x^2+2*x+4)

Use the difference of cubes formula, a^3 - b^3 = (a - b)(a^2 + ab + b^2) with a = x, b = 2.