Algebraic Identities Math Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

hard
Factor x3+8x^3 + 8 using the sum of cubes identity.

Solution

  1. 1
    x3+8=x3+23=(x+2)(x2−2x+4)x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4).
  2. 2
    Check: (x+2)(x2−2x+4)=x3−2x2+4x+2x2−4x+8=x3+8(x+2)(x^2-2x+4) = x^3 - 2x^2 + 4x + 2x^2 - 4x + 8 = x^3 + 8 ✓

Answer

(x+2)(x2−2x+4)(x + 2)(x^2 - 2x + 4)
The sum of cubes: a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2). The second factor is not factorable over the reals. Remember: 'Same sign, Opposite sign, Always Positive' (SOAP) for the signs.

About Algebraic Identities

Algebraic identities are equalities true for all permitted values of their variables.

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More Algebraic Identities Examples

Example 1 easy

Expand [formula] using the identity [formula].

Example 2 medium

Compute [formula] mentally using [formula] with [formula].

Example 3 easy

Expand [formula].

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