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Thread: Find the value of M

  1. #1
    Junior Member Dragon's Avatar
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    Find the value of M

    If m+n=3 and m^2+N^2=6 find the numerical value for m^3+N^3
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  2. #2
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    Quote Originally Posted by Dragon View Post
    If m+n=3 and m^2+N^2=6 find the numerical value for m^3+N^3
    You are told that,
    $\displaystyle n+m=3$
    Thus, squaring
    $\displaystyle (n+m)^2=9$
    $\displaystyle n^2+2nm+m^2=9$
    Thus,
    $\displaystyle 2mn+6=9$
    Thus,
    $\displaystyle 2 mn=3 \,$
    $\displaystyle mn=1.5$.

    Thus,
    $\displaystyle n^3+m^3=(n+m)(n^2-nm+m^2)=(3)(6-1.5)$
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  3. #3
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    Hello, Dragon!

    Another approach . . .


    If $\displaystyle m+n\,=\,3$ and $\displaystyle m^2+n^2\,=\,6$,
    find the numerical value for $\displaystyle m^3+n^3$

    We are given: .$\displaystyle \begin{array}{cc}(1)\\(2)\end{array} \begin{array}{cc}m + n \:=\:3 \\ m^2+n^2\:=\:6\end{array}$

    Square (1): .$\displaystyle (m + n)^2\:=\:3^2$

    . . . . . . . . . $\displaystyle \underbrace{m^2 + n^2} + 2mn\:=\:9$
    . . . . . . . . . - - $\displaystyle \downarrow$
    Substitute (2): .$\displaystyle 6 + 2mn \:=\:9\quad\Rightarrow\quad mn \,=\,\frac{3}{2}\quad(3)$


    Cube (1): .$\displaystyle (m+n)^3\:=\:3^3\quad\Rightarrow\quad m^3 + 3m^2n + 3mn^2 + n^3\:=\:27$

    . . . . . . . $\displaystyle \text{and we have: }\;\;m^3 + n^3 + 3\underbrace{mn}\underbrace{(m + n)}\:=\:27$
    . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle \downarrow\quad\;\downarrow$
    Substitute (3) and (1): .$\displaystyle m^3 + n^3 + 3\left(\frac{3}{2}\right)(3)\:=\:27$

    Therefore: .$\displaystyle m^3+n^3\:=\:\frac{27}{2}$

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