Simplify the given expression: \[{\left( {2x - 5y} \right)^3} - {\left( {2x + 5y} \right)^3}\]
A.\[120{x^2}{y^2} + 250{y^3}\]
B.\[ - 120{x^2}y + 250{y^3}\]
C.\[120{x^2}y + 250{y^3}\]
D.\[ - 120{x^2}y - 250{y^3}\]

Hint: In this question, we have given two terms, where one term is subtracted from another term. Each term has power three. Firstly, both terms (or expressions) write in their expanded form. After that, expanded form of respective expression, re-arrange the terms with their respective variable has same power. After solving (means respective terms will add their respective term or subtract), we get our answer. For this, we have following identities.
\[\left( i \right){\text{ }}{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)\]
\[\left( {ii} \right){\text{ }}{\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)\]

Complete step-by-step answer:
Step 1: Given \[{\left( {2x - 5y} \right)^3} - {\left( {2x + 5y} \right)^3} \ldots \ldots ......\left( A \right)\]
Firstly, expand \[{\left( {2x - 5y} \right)^3}\] with the help of \[{\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)\] identity.
Comparing \[{\left( {2x - 5y} \right)^3}\] with \[{\left( {a - b} \right)^3}\], we get the values:
\[a = 2x\] and value of \[b = 5y\].
After using the identity, we get:
\[{\left( {2x - 5y} \right)^3} = {\left( {2x} \right)^3} - {\left( {5y} \right)^3} - 3 \times 2x \times 5y \times \left( {2x - 5y} \right)\]
The cube of \[2x\] is \[8{x^3}\] and the cube of \[5y\] is \[125{y^3}\].
This implies,
\[{\left( {2x - 5y} \right)^3} = 8{x^3} - 125{y^3} - 30xy\left( {2x - 5y} \right)\]
Now, \[30xy\] multiplied with both terms, \[2x\] and \[5y\], i.e.,
\[30xy \times 2x = 60{x^2}y\] and \[30xy \times 5y = 150x{y^2}\]
So, we get:
\[{\left( {2x - 5y} \right)^3} = 8{x^3} - 125{y^3} - 60{x^2}y + 150x{y^2}\]
Step 2: \[{\left( {2x - 5y} \right)^3}\] can be expanded with the help of identity: \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)\]
Comparing \[{\left( {2x + 5y} \right)^3}\] with \[{\left( {a + b} \right)^3}\], we get the values:
\[a = 2x\] and value of \[b = 5y\].
After using the identity, we get:
\[{\left( {2x + 5y} \right)^3} = {\left( {2x} \right)^3} + {\left( {5y} \right)^3} + 3 \times 2x \times 5y \times \left( {2x + 5y} \right)\]
The cube of \[2x\] is \[8{x^3}\] and the cube of \[5y\] is \[125{y^3}\].
This implies,
\[{\left( {2x + 5y} \right)^3} = 8{x^3} + 125{y^3} + 30xy\left( {2x + 5y} \right)\]
After solving, we get:
$\Rightarrow$ \[{\left( {2x + 5y} \right)^3} = 8{x^3} + 125{y^3} + 60{x^2}y + 150x{y^2}\]
Put the value of \[{\left( {2x + 5y} \right)^3}\] and \[{\left( {2x - 5y} \right)^3}\] in equation (A), we get:
\[{\left( {2x - 5y} \right)^3} - {\left( {2x + 5y} \right)^3} = \left( {8{x^3} - 125{y^3} - 60{x^2}y - 150x{y^2}} \right) - \left( {8{x^3} + 125{y^3} + 60{x^2}y + 150x{y^2}} \right)\]
\[ \Rightarrow 8{x^3} - 125{y^3} - 60{x^2}y + 150x{y^2} - 8{x^3} - 125{y^3} - 60{x^2}y - 150x{y^2}\]
After cancellation of respective terms, we get:
\[ = - 125{y^3} - 60{x^2}y - 125{y^3} - 60{x^2}y\]
Re-arranging the terms, we get:
\[ \Rightarrow - 125{y^3} - 125{y^3} - 60{x^2}y - 60{x^2}y\]
Here, \[ - 125{y^3} - 125{y^3}\], we get \[ - 250{y^3}\] and \[ - 60{x^2}y - 60{x^2}y\], we get \[ - 120{x^2}y\]
After solving the respective terms, we get:
\[{\left( {2x - 5y} \right)^3} - {\left( {2x + 5y} \right)^3} = - 120{x^2}y - 250{y^3}\]
Therefore, option (D) is correct.

Note: A polynomial of one term is called a monomial. A polynomial of two terms is called a binomial. A polynomial of three terms is called a trinomial. A polynomial of degree three is called a cubic polynomial. Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial has no zero and every real number is a zero of the zero polynomial. A polynomial \[p\left( x \right)\] in one variable \[x\] is an algebraic expression in \[x\] of the form:
\[p\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........{a_1}x + {a_0}x\]