Courses for Kids
Free study material
Offline Centres
Store Icon

Find the product of\[\left( m+n \right)\left( {{m}^{2}}-mn+{{n}^{2}} \right)\].

Last updated date: 23rd Jul 2024
Total views: 451.5k
Views today: 13.51k
451.5k+ views
Hint:Given is an algebraic expression. Open the brackets and simplify the expression. You will get the product of the expression.

Complete step-by-step answer:
The given expression is an algebraic expression built up from integer constants, variables and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent) that is a rational number.
Given, \[\left( m+n \right)\left( {{m}^{2}}-mn+{{n}^{2}} \right)\].
Opening the brackets and simplifying the expression,
  & \left( m+n \right)\left( {{m}^{2}}-mn+{{n}^{2}} \right)=m\left( {{m}^{2}}-mn+{{n}^{2}} \right)+n\left( {{m}^{2}}-mn+{{n}^{2}} \right) \\
 & =\left[ m\times {{m}^{2}}-m\times \left( mn \right)+m\times {{n}^{2}} \right]+\left[ n\times {{m}^{2}}-n\times \left( mn \right)+n\times {{n}^{2}} \right] \\
 & =\left[ {{m}^{3}}-{{m}^{2}}n+m{{n}^{2}} \right]+\left[ n{{m}^{2}}-m{{n}^{2}}+{{n}^{3}} \right] \\
Opening the brackets and simplifying it,
Cancel out similar terms \[m{{n}^{2}}\]and \[{{m}^{2}}n\].
\[\therefore \left( m+n \right)\left( {{m}^{2}}-mn+{{n}^{2}} \right)={{m}^{3}}+{{n}^{3}}\]
Hence, we have found the product of the algebraic expression.

When removing the brackets and multiplying them be careful as not to mix up the sign and the terms m and n. If changing a sign or mistakenly putting n instead of m would change the entire answer.
Suppose we have an expression\[{{a}^{3}}+{{b}^{3}}\].
The expanded from of\[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\].
which is similar to our question\[\left( m+n \right)\left( {{m}^{2}}-mn+{{n}^{2}} \right)\].
So without opening brackets and expanding it we guess the answer as\[{{m}^{3}}+{{n}^{3}}\].