Question

How do you simplify `sqrt ((4a^3 )/( 27b^3))`?

Answer 1

`sqrt((4a^3)/(27b^3))=(2a)/(3b)sqrt(a/(3b))`

Explanation:

`sqrt((4a^3)/(27b^3))`

= `sqrt((2xx2xxaxxaxxa)/(3xx3xx3xxbxxbxxb))`

= `sqrt((ul(2xx2)xxul(axxa)xxa)/(ul(3xx3)xx3xxul(bxxb)xxb))`

= `(2a)/(3b)sqrt(a/(3xxb))`

= `(2a)/(3b)sqrt(a/(3b))`

Answer 2

Just another way of writing the same thing as the other solution:

`" "color(green)((2asqrt(3ab))/(3b^2))`

Explanation:

Given:`" "sqrt((4a^3)/(27b^3))`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Concept of approach")`

You look for numbers that are squared. Take them outside the square root and 'get rid' of the square.

Example: Suppose we had `sqrt(12)`

Choosing the factors of `3xx4=12` we write it as:

`sqrt(3xx4)` but 4 is `2^2` giving `sqrt(3xx2^2)`

Take the 2 out side the square root giving

`" "2sqrt(3)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving your question")`

`color(brown)("Mathematicians do not like square roots in the denominator so we will 'get rid' of it")` `color(brown)("later.")`

`" Write as: "(sqrt(4a^3))/(sqrt(27b^3))`
'.......................................................................................
we know that
`27" "=" "3xx9" " =" "3xx3^2" and "b^3" "=" "b^2xxb`
`4" "=2^2" and " a^3" "=a^2xxa`
'.........................................................................................

`" Write as "sqrt(2^2xxa^2xxa)/sqrt(3^2xx3xxb^2xxb)`

Take the squared values outside the square roots giving

`" "color(blue)((2a sqrt(a))/(3bsqrt(3b)))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Now to 'get rid' of the square root in the denominator")`

Multiply by 1 and you do not change the value or how it looks.
1 can come in many forms: `1/1"; "(-1)/(-1)"; "2/2"; "sqrt(3b)/sqrt(3b)`

So we can multiply by 1 and not change the inherent value but we can change the way it looks.

`color(brown)("Multiply by 1 but in the form of "1=sqrt(3b)/sqrt(3b))`

`(2a sqrt(a))/(3bsqrt(3b))xxsqrt(3b)/sqrt(3b)" "=" "(2asqrt(a)sqrt(3b))/(3bsqrt(3b)sqrt(3b))`

But `sqrt(3b)xxsqrt(3b)=3b` giving:

`(2asqrt(a)sqrt(3b))/(3bxxb)" "=" "(2asqrt(a)sqrt(3b))/(3b^2)`

But `sqrt(a)sqrt(3b) = sqrt(3ab)`

`color(brown)("If you can take things out of square roots you can put things back in!")`

`" "color(green)((2asqrt(3ab))/(3b^2))`