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Math Help - Factoring Sum of Cube

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    Senior Member vaironxxrd's Avatar
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    Factoring Sum of Cube

    I was given the formula a^3+b^3=&(a+b)(a^2-ab+b^2) For cubes.

    But this problem is kind of weird for me since I don't know if I should find the cube of x^6

    Problem: x^6+125
    Attempt: (x+5)(x^2+5x+25) Distributed: 5x^3+25x+125
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    Re: Factoring Sum of Cube

    Quote Originally Posted by vaironxxrd View Post
    I was given the formula a^3+b^3=&(a+b)(a^2-ab+b^2) For cubes.

    But this problem is kind of weird for me since I don't know if I should find the cube of x^6

    Problem: x^6+125
    Attempt: (x+5)(x^2+5x+25) Distributed: 5x^3+25x+125
    It's because \displaystyle x^6 + 125 = \left(x^2\right)^3 + 5^3, not \displaystyle x^3 + 5^3...
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    Re: Factoring Sum of Cube

    Quote Originally Posted by vaironxxrd View Post
    I was given the formula a^3+b^3=&(a+b)(a^2-ab+b^2) For cubes.
    Problem: x^6+125
    x^6+125=(x^2+5)(x^4-5x^2+25)
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  4. #4
    Senior Member vaironxxrd's Avatar
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    Re: Factoring Sum of Cube

    Quote Originally Posted by Prove It View Post
    It's because \displaystyle x^6 + 125 = \left(x^2\right)^3 + 5^3, not \displaystyle x^3 + 5^3...
    Got it I saw that it's also a way to do it but I really Din't know I could use it here.
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