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motikmotik
3 years ago
6

Could someone help me, please?

Mathematics
1 answer:
k0ka [10]3 years ago
3 0

We can rewrite the expression under the radical as

\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4

then taking the fourth root, we get

\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|

Why the absolute value? It's for the same reason that

\sqrt{x^2}=|x|

since both (-x)^2 and x^2 return the same number x^2, and |x| 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|

The absolute values disappear on all but the b term because all of \dfrac32, a^2 and c^4 are positive, while b^3 could potentially be negative. So we end up with

\dfrac32a^2\left|b^3\right|c^4=\dfrac32a^2|b|^3c^4

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