Answer:
Step-by-step explanation:
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We can rewrite the expression under the radical as
![\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4](https://tex.z-dn.net/?f=%5Cdfrac%7B81%7D%7B16%7Da%5E8b%5E%7B12%7Dc%5E%7B16%7D%3D%5Cleft%28%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%29%5E4)
then taking the fourth root, we get
![\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cleft%28%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%29%5E4%7D%3D%5Cleft%7C%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%7C)
Why the absolute value? It's for the same reason that
![\sqrt{x^2}=|x|](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2%7D%3D%7Cx%7C)
since both
and
return the same number
, and
captures both possibilities. From here, we have
![\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|](https://tex.z-dn.net/?f=%5Cleft%7C%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%7C%3D%5Cleft%7C%5Cdfrac32%5Cright%7C%5Cleft%7Ca%5E2%5Cright%7C%5Cleft%7Cb%5E3%5Cright%7C%5Cleft%7Cc%5E4%5Cright%7C)
The absolute values disappear on all but the
term because all of
,
and
are positive, while
could potentially be negative. So we end up with
![\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4](https://tex.z-dn.net/?f=%5Cdfrac32a%5E2%5Cleft%7Cb%5E3%5Cright%7Cc%5E4%3D%5Cdfrac32a%5E2%7Cb%7C%5E3c%5E4)
Triangles are the easiest shape to make a hexagon out of. You can do this by lining up 6 triangles where each base of the triangle lines up.