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Math Help - sum and differences? 9th grade?

  1. #1
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    sum and differences? 9th grade?

    Use these patterns to factor polynomials that are sums and differences of cubes.
    a^3 + b^3 = (a+b)(a^2-ab+b^2)
    a^3-b^3=(a-b)(a^2+ab+b^2)

    a)x+8
    b)8x^3+27
    c)64-y^3
    d)2u^3 + 16V^3
    e)d^5 + f^3
    f) w^6 -1

    2. if the sum of two numbers is 6 and the sum of their cubes is 18, what's the sum of their squares?

    3. use synthetic division to find (x^2 + 4) / (s - 2i)


    Please show work! Thank you so so much!!!!
    Give me the answers too.
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  2. #2
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    Quote Originally Posted by tecktonikk View Post
    Use these patterns to factor polynomials that are sums and differences of cubes.
    a^3 + b^3 = (a+b)(a^2-ab+b^2)
    a^3-b^3=(a-b)(a^2+ab+b^2)

    a)x+8 \textcolor{red}{(\sqrt[3]{x})^2 + 2^3}

    b)8x^3+27 \textcolor{red}{(2x)^3 + 3^3}

    c)64-y^3 \textcolor{red}{4^3 - y^3}

    d)2u^3 + 16V^3 \textcolor{red}{2[u^3 + (2v)^3]}

    e)d^5 + f^3 \textcolor{red}{(d^{\frac{5}{3}})^3 + f^3}

    f) w^6 -1 \textcolor{red}{(w^2)^3 - 1^3}

    2. if the sum of two numbers is 6 and the sum of their cubes is 18, what's the sum of their squares?

    \textcolor{red}{x+y = 6}

    \textcolor{red}{x^3+y^3 = 18}

    also, note that \textcolor{red}{(x+y)^2 = 36}

    3. use synthetic division to find (x^2 + 4) / (s - 2i)
    Code:
    2i] 1 ..... 0 ...... 4
    
    ----------------------
    all set up ... you do the work .
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  3. #3
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    Hello, tecktonikk!

    2. If the sum of two numbers is 6 and the sum of their cubes is 18,
    what is the sum of their squares?
    Let the two numbers be x and y.

    We have: . \begin{array}{cccc}x+y &=& 6 & {\color{blue}[1]} \\ x^3+y^3 &=& 18 & {\color{blue}[2]} \end{array}


    Cube equation [1]: . (x+y)^3 \:=\:6^3 \quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:216

    \text{We have: }\;\underbrace{x^3 + y^3}_{18} + \; 3xy\underbrace{(x+y)}_6 \;=\;216 \quad\Rightarrow\quad 18 \;+\; 18xy \:=\:216 \quad\Rightarrow\quad xy \:=\:11


    Square equation [1]: . (x+y)^2 \:=\:6^2 \quad\Rightarrow\quad x^2 + 2xy + y^2 \:=\:36

    \text{We have: }\;x^2 + y^2 + 2\underbrace{(xy)}_{11} \:=\:36 \quad\Rightarrow\quad x^2 + y^2 + 22 \:=\:36 \quad\Rightarrow\quad\boxed{x^2+y^2 \:=\:14}

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