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Math Help - Problem #3 - Cubic Expression

  1. #1
    Forum Admin topsquark's Avatar
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    Problem #3 - Cubic Expression

    Prove the following:

    \sqrt[3]{9 + \sqrt{80}} + \sqrt[3]{9 - \sqrt{80}} = 3

    -Dan
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  2. #2
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    re: Problem #3 - Cubic Expression

    easy let cuberoot(9+sqrt(80))+ cuberoot(9-sqrt(80))=x
    now cube on both l.h.s and r.h.s
    apply (a+b)^(3)=a^3+b^3+3ab(a+b)
    now
    x^(3)=9+sqrt(80)+9-sqrt(80)+3*cuberoot(9+sqrt(80))*cuberoot(9-sqrt(80))*(x)
    now (9-sqrt(80))=1/(9+sqrt(80))
    therefore the two cuberoot terms would be cancelled
    now
    x^(3)=18-3x
    x^(3)+3x-18=0
    now apply factor theorem take (x-3) and divide this to the above cubic eqn
    we thus get x=3 as root
    substitute the value of x and u will get the result
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  3. #3
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    re: Problem #3 - Cubic Expression

    Quote Originally Posted by prasum View Post
    easy <snip>
    Please wrap these answers in spoiler flags

    Spoiler:
    like this

    so that people don't immediately see your answer when viewing the thread.

    Others would like a chance to do the problem themselves.

    It's the button that looks like an eye on the advanced editing toolbar.
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  4. #4
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    re: Problem #3 - Cubic Expression

    Hello, topsquark!

    Prove: . \sqrt[3]{9 + \sqrt{80}} + \sqrt[3]{9 - \sqrt{80}} \;=\: 3

    Spoiler:

    Let: . A \,=\,9 + \sqrt{80},\;B \,=\,9-\sqrt{80}

    Note that: . A + B \:=\: (9+\sqrt{80}) + (9-\sqrt{80}) \:=\:18
    . . . . . . . . . . . AB \:=\:(9+\sqrt{80})(9 - \sqrt{80}) \:=\:1


    We have: . X \:=\:A^{\frac{1}{3}} + B^{\frac{1}{3}}

    Cube both sides:
    . . X^3 \;=\;\left(A^{\frac{1}{3}} + B^{\frac{1}{3}}\right)^3

    . . X^3 \;=\;A + 3A^{\frac{2}{3}}B^{\frac{1}{3}} + 3A^{\frac{1}{3}}B^{\frac{2}{3}} + B

    . . X^3 \;=\;(A + B) + 3A^{\frac{1}{3}}B^{\frac{1}{3}}\left(A^{\frac{1}{3  }} + B^{\frac{1}{3}}\right)

    . . X^3 \;=\;\underbrace{(A+B)}_{18} + 3\underbrace{(AB)^{\frac{1}{3}}}_{1^\frac{1}{3}}} \underbrace{\left(A^{\frac{1}{3}} + B^{\frac{1}{3}}\right)}_3

    . . X^3 \;=\;18 + 3(1)(3)

    . . X^3 \;=\;27

    Therefore: . X \;=\;3

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  5. #5
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    re: Problem #3 - Cubic Expression

    .
    Spoiler:
    easy let cuberoot(9+sqrt(80))+ cuberoot(9-sqrt(80))=x
    now cube on both l.h.s and r.h.s
    apply (a+b)^(3)=a^3+b^3+3ab(a+b)
    now
    x^(3)=9+sqrt(80)+9-sqrt(80)+3*cuberoot(9+sqrt(80))*cuberoot(9-sqrt(80))*(x)
    now (9-sqrt(80))=1/(9+sqrt(80))
    therefore the two cuberoot terms would be cancelled
    now
    x^(3)=18-3x
    x^(3)+3x-18=0
    now apply factor theorem take (x-3) and divide this to the above cubic eqn
    we thus get x=3 as root
    substitute the value of x and u will get the result
    Thanks from topsquark
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  6. #6
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    re: Problem #3 - Cubic Expression

    I must admit that it took me a while to understand prasum's answer, but I eventually got there. Possibly I drank too much wine with dinner last night because I still don't understand soroban's answer.

    Prove $\sqrt[3]{9 + \sqrt{80}} + \sqrt[3]{9 - \sqrt{80}} = 3.$

    Eventually prasum gets to $x^3 = 18 - 3x$ but he means $x^3 = 18 + 3x.$

    $x^3 = 18 + 3x \implies x^3 - 3x - 18 = 0 \implies (x - 3)(x^2 + 3x + 6) = 0 \implies x = 3\ if\ x \in \mathbb R.$

    So the typo had me stuck for a while.

    Soroban defines $A = 9 + \sqrt{80}\ and\ B = 9 - \sqrt{80}.$ In his notation he must prove $\sqrt[3]{A} + \sqrt[3]{B} = 3.$ I'm fine with that obviously.

    He asserts that $A + B = 18.$ Even I can deduce that from his definitions.

    He asserts that $\sqrt[3]{AB} = 1.$ I can also deduce that from his definitions.

    And he asserts that $\sqrt[3]{A} + \sqrt[3]{B} = 3.$ But he can't assert that because that is what is to be proved, or am I missing something?
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  7. #7
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    re: Problem #3 - Cubic Expression

    yes the secondlast step of Sorobans answer must have (A^(1/3)+B^(1/3))=X not 3
    then X would be the required answer
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  8. #8
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    Re: Problem #3 - Cubic Expression

    Congrats to everyone this week!

    -Dan
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