pre calc problem: simplify 12 ³?(648x^4)+ 3³?(81x^4)

[harris]

pre calc problem: simplify 12 ³?(648x^4)+ 3³?(81x^4)

simplify 12 ³?(648x^4)+ 3³?(81x^4)

 

[Loren]

Re: pre calc problem: simplify 12 ³?(648x^4)+ 3³?(81x^4)

648 = 2*2*2*3*3*3*3, so \(\displaystyle \sqrt[3]{648} = 2*3\sqrt[3]{3}=6\sqrt[3]{3}\).
\(\displaystyle \sqrt[3]{x^4}=x\sqrt[3]{x}\).
Now, you finish it.

 

[soroban]

Re: pre calc problem: simplify 12 ³?(648x^4)+ 3³?(81x^4)

Hello, harris!

Another approach . . .

Simplify: .\(\displaystyle 12\sqrt[3]{648x^4} + 3\sqrt[3]{81x^4}\)

We are looking for factors which are cubes . . .


\(\displaystyle \text{The first radical is: }\:\sqrt[3]{648x^4} \;=\;\sqrt[3]{216\cdot 3\cdot x^3\cdot x} \;=\;\sqrt[3]{216x^3\cdot 3x} \;=\;\sqrt[3]{216x^3}\cdot\sqrt[3]{3x} \;=\;6x\sqrt[3]{3x}\)

\(\displaystyle \text{The second radical is: }\:\sqrt[3]{81x^4} \;=\;\sqrt[3]{27\cdot3\cdot x^3\cdot x} \;=\;\sqrt[3]{27x^3\cdot 3x} \;=\;\sqrt[3]{27x^3}\cdot\sqrt[3]{3x}\;=\;3x\sqrt[3]{3x}\)


\(\displaystyle \text{The problem becomes: }\:12\cdot\!\left( 6x\sqrt[3]{3x}\right) \;+ \;3\cdot\!\left(3x\sqrt[3]{3x}\right) \;=\;72x\sqrt[3]{3x} \;+\; 9x\sqrt[3]{3x} \quad\hdots\quad\text{ Got it?}\)