# Roots

• Nov 16th 2009, 11:29 AM
james_bond
Roots
Show that $(4-x)\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4}=4$ if $a=1+\sqrt5$.
• Nov 17th 2009, 01:22 AM
Opalg
Quote:

Originally Posted by james_bond
Show that $(4-{\color{red}a})\cdot\sqrt{2+a}\cdot\sqrt[3]{a}\cdot\sqrt[6]{3a+4}=4$ if $a=1+\sqrt5$.

If $a=1+\sqrt5$ then $a^2 = 6+2\sqrt5$ and $3a+4 = 7+3\sqrt5$. Multiply these together to check that $a^2(3a+4) = 72+32\sqrt5$.

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

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