Question

Simplify the expression.

Cubed root of 27(x-2)^3.

Answer 1

\[\sqrt[3]{27(x-2)^3}= \sqrt[3]{3^3*(x-2)^3}\]
in general cube root(y^3) = y
so

\[\sqrt[3]{3^3(x-2)^3}=3*(x-2)\]

Answer 2

So it would not be 27(x-2) because you can simplify the 27, correct?

Answer 3

yes because you use 3^3

Answer 4

Does the cubed number and the power number cancel each other out?

Answer 5

yes, cube root means a power of 1/3. so when you have cube root of 3^3, you have 3^(3/3) = 3^1 = 3

Answer 6

Sorry I suck at math....

Answer 7

The answer is actually:\[3x-6\]:)

Answer 8

How do you get 3x-6?

Answer 9

\[3(x-2)=3(x)+3(-2)=3x-6\] I used distribution which is basically\[a(b+c)=ab+ac\]

Answer 10

Get it or still having trouble with it?

Answer 11

Not using the distributive property. It's the cubed root to the third power. Unless I'm throughly confused, which is probably the case....

Answer 12

Ok, the thing is that I am using the distributive to the answer 3(x-2) not the whole thing. This is the whola thing:\[\sqrt[3]{27(x-2)^{3}}=\sqrt[3]{27}\sqrt[3]{(x-2)^{3}}=3(x-2)=3(x)-3(2)=3x-6\]Remember that a cubed root is distributive can be broken down in multiplications and that a cubed power cancels a squared root.

Answer 13

ok...thank you ....finally got it

Answer 14

No prob