
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.\[ - 120{x^2}y - 250{y^3}\]
Answer
564.6k+ views
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:
\[p\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........{a_1}x + {a_0}x\]
\[\left( i \right){\text{ }}{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)\]
\[\left( {ii} \right){\text{ }}{\left( {a - b} \right)^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)\]
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:
\[p\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........{a_1}x + {a_0}x\]
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

Trending doubts
Which are the Top 10 Largest States of India?

What is the theme or message of the poem The road not class 9 english CBSE

Define development

Distinguish between population growth and population class 9 social science CBSE

Explain the importance of pH in everyday life class 9 chemistry CBSE

The winter rain in Chennai is caused by A SouthWest class 9 social science CBSE

