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Thread: Algberic question

  1. #1
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    Algberic question

    If x^a*x^b*x^c =1 then the value of a^3+b^3+c^3 is


    a)9

    b)abc

    c) a+b+c

    d) 3 abc
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Algberic question

    Given:

    $\displaystyle x^ax^bx^c = x^{(a+b+c)} = 1$

    then $\displaystyle a+b+c = 0$, and so $\displaystyle c= -(a+b)$. Hence:

    $\displaystyle a^3 + b^3 + c^3 = a^3 + b^3 - (a+b)^3 = a^3 + b^3 -(a^3 + 3a^2b + 3b^2c + b^3) = -3ab( a+b) = 3abc $
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  3. #3
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    Re: Algberic question

    Hello, sscgeek!

    $\displaystyle \text{Given: }\:x^ax^bx^c \:=\:1$

    $\displaystyle \text{Find the value of: }\:a^3+b^3+c^3$

    . . $\displaystyle (a)\;9 \qquad (b)\;abc \qquad (c)\;a+b+c \qquad (d)\;3 abc$

    We have: .$\displaystyle x^{a+b+c} \:=\:1 \quad\Rightarrow\quad a+b+c \:=\:0$

    Cube both sides: .$\displaystyle (a+b+c)^3 \:=\:0^3 $

    . . $\displaystyle a^3 + 3a^2b + 3ab^2 + b^3 + 3b^2c + 3bc^2 + 3a^2c + 3ac^2 + c^3 + 6abc \:=\:0$

    . . $\displaystyle a^3 + b^3 + c^3 + (3a^2b+2ab^2 + 3abc)$
    . . . . . . . . . . . . . . $\displaystyle + (3b^2c+3bc^2 + 3abc) + (3a^2c+3ac^2 {\color{blue}+\: 3abc}) {\color{blue}-\: 3abc} \:=\:0$

    . . $\displaystyle a^3+b^3+c^3 + 3ab\underbrace{(a+b+c)}_{\text{This is 0}} + \:3bc\underbrace{(a+b+c)}_{\text{This is 0}} + \:3ac\underbrace{(a+b+c)}_{\text{This is 0}} -\:3abc\:=\:0$

    . . . . . . . $\displaystyle a^3+b^3+c^3 - 3abc \;=\;0$

    . . . . . . . . . $\displaystyle a^3+b^3+c^3 \;=\;3abc\;\;\hdots\;\;\text{answer (d)}$
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