Question

HELP PLZ and explain:)

Answer 1

Given x, y, a, b are positive numbers

`x^(-2/3)=a^(-2)=>x^(-1/3)=\pm{a^(-1)}=>x^(-1/3)=a^(-1)........[x,a >0]`

`y^(2/3)=b^4=>y^(1/3)=\pm{b^2}=>y^(1/3)=b^2..........[y,b >0]`


`(xy)^(-1/3)=x^(-1/3)*y^(-1/3)`

`=x^(-1/3)*y^(-1/3)`

`=a^(-1)*b^(-2)`

`=(1)/(ab^2)`.

Answer 2

First, remember: taking the nth root of a number a is the same as raising that number to the 1/n power or:
`\sqrt[n]{a} = a^{ (1)/(n)}`.
>> That may sound super mathematical, but here's an example to help you understand it better:
Let's say you're trying to find the cube root of a number a:
`\sqrt[3]{a}` and a is a bizarre value like
`x^(120)`:

Since
`\sqrt[3]{a} = a^{ (1)/(3)}`, you would calculate:
`\sqrt[3]{a} = \sqrt[3]{x^(120)} = x^{120 * (1)/(3)}`. Remember that you multiply exponents when you're raising a power to a power!
Finally, simplify:
`x^{120 * (1)/(3)} = x^{(120)/(3)} = x^(40)`

----

Let's get back to the problem:
This is a problem you'll have to take step-by-step. Since you want to find
`(xy)^{- (1)/(3) }` in terms of a and b, you know you'll have to substitute a in for x and y in for b. That means you'll have to first isolate the variables x and y:

1) You're given
`x^{- (2)/(3)} = a^(-2)`. To solve this for x, you must take the
`- (2)/(3)` root of both sides or raise both sides to
`(1)/(-(2)/(3)) = - (3)/(2)`:

`x^{- (2)/(3)} = a^(-2) \\ x^{- (2)/(3)* - (3)/(2)} = a^{-2* - (3)/(2)} \\ x^(1) = a^(3) \\ x=a^(3)`

2) You're also given
`y^{ (2)/(3)} = b^(4)`. To solve for y, take the
`(2)/(3)` root of both sides, or raise both sides to
`(1)/( (2)/(3)) = (3)/(2)`:

`y^{ (2)/(3)} = b^(4) \\ y^{ (2)/(3)* (3)/(2) } = b^{4* (3)/(2)} \\ y^( 1) = b^(6) \\ y = b^(6)`

Now you have x and y in terms of a and b: x=a^{3} and y = b^{6}. Put it into the equation
`(xy)^{- (1)/(3)}`:

`(xy)^{- (1)/(3)} = (a^(3)b^(6))^{- (1)/(3)}`

To simplify, remember another rule about exponents:
`(ab)^(n) = a^(n) * b^(n)`. That means you can distribute the
`(-1)/(3)` power to both the
`a^(3)` and the
`b^(6)`:

`(a^(3)b^(6))^{- (1)/(3)} \\ = ( a^{3*- (1)/(3)} )(b^{6*- (1)/(3)}) \\ =a^(-1)b^(- 2)`


`a^(-1)b^(- 2)` is your final answer, unless you want to simplify it further using the negative exponents rule
`b^(-n)= (1)/( b^(n))`, meaning:

`a^(-1)b^(- 2) \\ = (1)/( a^(1)) * (1)/( b^(2)) \\ = (1)/(ab^(2))`

So,
`(1)/(ab^(2))` is your completely simplified final answer.

This might be confusing at first if you're not familiar with exponent rules, so let me know if you're confused. Hope this helps and feel free to ask questions!