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Thread: Roots

  1. #1
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    Roots

    Show that $\displaystyle (4-x)\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4}=4$ if $\displaystyle a=1+\sqrt5$.
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Show that $\displaystyle (4-{\color{red}a})\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4}=4$ if $\displaystyle a=1+\sqrt5$.
    If $\displaystyle a=1+\sqrt5$ then $\displaystyle a^2 = 6+2\sqrt5$ and $\displaystyle 3a+4 = 7+3\sqrt5$. Multiply these together to check that $\displaystyle a^2(3a+4) = 72+32\sqrt5$.

    Next, $\displaystyle (2+a)^3 = (3+\sqrt5)^3$. check that is also equal to $\displaystyle 72+32\sqrt5$. Therefore $\displaystyle \sqrt{2+a} = \bigl(a^2(3a+4)\bigr)^{1/6} = \sqrt[3]{a}\cdot\sqrt[6]{3a+4}$.

    Therefore $\displaystyle (4-a)\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4} = (4-a)(2+a) = (3-\sqrt5)(3+\sqrt5) = 9-5=4$.
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