How do you factor x^3+8 using the sum-of-two cubes?

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How do you factor x^3+8 using the sum-of-two cubes?

2006-09-02 07:18:22 · 5 answers · asked by teena 1 in Science & Mathematics ➔ Mathematics

5 answers

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

2006-09-02 07:21:30 · answer #1 · answered by raj 7 · 0⤊ 0⤋

The formula to factor the sum of two cubes is x^3 + y^3 = (x + y) * (x^2 - xy + y^2). In this case, we have x^3 + 2^3, so that's (x + 2) * (x^2 - 2x + 4).

2006-09-02 07:24:02 · answer #2 · answered by DavidK93 7 · 0⤊ 0⤋

The factor of x^3+8 is (x+2) (x^2-2x+4). Trust me!

2006-09-02 07:35:44 · answer #3 · answered by Ijsselor 1 · 0⤊ 0⤋

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

2006-09-02 08:38:10 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋

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

2006-09-03 03:08:12 · answer #5 · answered by pavan kumar 1 · 0⤊ 0⤋