How do you simplify ${{2{x^3}}} \times {{{y^{-3}}}} .{{2{x^{-1}}}}{y^3}$ and write it using only positive exponents ?

Hint: To simplify this question , we need to solve it step by step . We should know that the exponent having -1 as power is considered to be $ $ $ \dfrac{1}{x} $ . So , by rewriting all the powers having negative exponents can be rewritten by using only positive exponents . Here we have two variables having negative exponents , we will rewrite it similarly and try to simplify by cancelling out the common factors to get the required result .

Complete step-by-step answer:
In order to solve the expression , write it the way it is given and then arrange according to the positive exponents by making them the denominator .
As per our given Question ${{2{x^3}}} \times {{{y^{-3}}}} .{{2{x^{-1}}}}{y^3}$ ,
The exponents having $ {x^{ - 1}} $ , -1 as power is considered to be $ $ $ \dfrac{1}{x} $ .
Rewrite the above accordingly as given in the expression , we have =
We can further simplify it by cancelling the common factors as –
 $\Rightarrow \dfrac{{2{x^3}}}{{{y^3}}} \times \dfrac{{2{y^3}}}{x} $
The variable $ {y^3} $ can be cancelled out as it is both in denominator and numerator making it is unity .
 $\Rightarrow \dfrac{{2{x^3}}}{1} \times \dfrac{{2 \times 1}}{x} $
Now , further $ {x^3} $ in the numerator can be cancelled by $ x $ raised to the power 1 by just leaving $ {x^2} $ in the numerator as –
 $\Rightarrow 2{x^2} \times 2 $
Further on simplifying we will get ,
 $ 4{x^2} $ which is having a positive exponent .
Therefore, the final solution to this question is $ 4{x^2} $ .
So, the correct answer is “ $ 4{x^2} $ ”.

Note: Always remember you can perform calculations only between like terms .
Always try to work on the positive exponents .
The expression is having hidden multiplication when written altogether .
Do not forget to verify the exponents solved correctly .
Always try to cancel out the similar terms for the solution of simplification .