Question

Can you help me figure out how to find the original radican ??? I have no clue how to do so

Answer 1

So we have:

`-3a^5b^2\sqrt[3]{a^2c}`

And we want to knowthe original before simplification, that is, before evaluating the interior part of the root.

So, we need to figure a way to put the part outside of the root back in.

Taking the cubic root of a number is the same as dividing its exponent by 3, because:

`\sqrt[3]{a^n}=a^{(n)/(3)}`

So, thinking in the other direction, we need to multiply the exponents by 3 before taking it back to the inside of the cubic root:

`a^k=a^{(3k)/(3)}=\sqrt[3]{a^(3k)}`

So, the b part have a 2 in the exponent, so we can multiply it by 3 to get 6:

`\begin{gathered} b^2=b^{(3\cdot2)/(3)}=\sqrt[3]{b^(3\cdot2)}=\sqrt[3]{b^6} \\ -3a^5b^2\sqrt[3]{a^2c}=-3a^5\sqrt[3]{b^6}\sqrt[3]{a^2b^(3\cdot2)c}=-3a^5\sqrt[3]{a^2b^6c} \end{gathered}`

The a part have a 5 in the exponent, so we will get 15:

`\begin{gathered} a^5=a^{(3\cdot5)/(3)}=\sqrt[3]{a^(3\cdot5)}=\sqrt[3]{a^(15)} \\ -3a^5\sqrt[3]{a^2b^6c}=-3\sqrt[3]{a^(15)}\sqrt[3]{a^2^{}b^6c}=-3\sqrt[3]{a^2a^(15)b^6c} \end{gathered}`

Now, since we have a² and a¹⁵, we can add their exponents:

`\begin{gathered} a^2a^(15)=a^(17) \\ -3\sqrt[3]{a^2a^(15)b^6c}=-3^{}\sqrt[3]{a^(17)b^6c} \end{gathered}`

Now, the -3 have an exponent of 1, so:

`\begin{gathered} -3=(-3)^1=(-3)^{(3\cdot1)/(3)}=\sqrt[3]{(-3)^(3\cdot1)}=\sqrt[3]{(-3)^3}=\sqrt[3]{-27} \\ -3^{}\sqrt[3]{a^(17)b^6c}=\sqrt[3]{-27}^{}\sqrt[3]{a^(17)b^6c}=^{}\sqrt[3]{-27a^(17)b^6c} \end{gathered}`

Thus, we have, in the end:

`^{}\sqrt[3]{-27a^(17)b^6c}=-3a^5b^2^{}\sqrt[3]{a^2^{}c}`