hi. I'm expected to solve (27ab^3)^1/3 times the sq. root of (there is a three in the nook of the square root) 5a^4b. I follow it until it reduces to the square root of (still with the three in the nook), 3^3a^3b^3 times 5 a^2 b. Where does the three on the first a randomly come from? how does the a randomly get squared? Thanks for your help!!

Question

hi. I'm expected to solve (27ab^3)^1/3 times the sq. root of (there is a three in the nook of the square root) 5a^4b. I follow it until it reduces to the square root of (still with the three in the nook), 3^3a^3b^3 times 5 a^2 b. Where does the three on the first a randomly come from? how does the a randomly get squared? Thanks for your help!!

Answer 1

if your first factor had been

(27(ab)^3)^1/3 that would become
3ab

Did you forget a bracket?

Answer 2

Hello! Let me break down the steps to help clarify where the exponents and additional terms come from.

The given expression is:

(27ab^3)^(1/3) * √(3) * 5a^4b

To simplify this expression, we can start by evaluating the cube root of (27ab^3).

Step 1: Cube root of (27ab^3)
The cube root of a number is equivalent to raising it to the power of 1/3.

(27ab^3)^(1/3)

To simplify this, you can apply the exponent to each term:

27^(1/3) * a^(1/3) * b^[(3 * 1/3)]
= 3 * a^(1/3) * b^1
= 3ab

Now, let's substitute this simplified expression back into the original equation.

Step 2: Substitute the simplified expression back into the original equation
(3ab) * √(3) * 5a^4b

Now, we need to simplify the term a^4b.

Step 3: Simplify the term a^4b
a^4b can be simplified by applying the laws of exponents. When we multiply variables with the same base, we add their exponents.

a^4b = a^(3+1) * b^1
= a^3 * a^1 * b
= a^3b

Substituting this simplified term back into the equation, we get:

(3ab) * √(3) * 5a^3b

Therefore, the correct simplified expression is the square root of (3) times 3^3a^3b^3 times 5a^2b. There is no random three or squared 'a'; they come from applying the laws of exponents during simplification.