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

Answer

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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,

\[\begin{align}

& \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] \\

\end{align}\]

Opening the brackets and simplifying it,

\[={{m}^{3}}-{{m}^{2}}n+m{{n}^{2}}+n{{m}^{2}}-m{{n}^{2}}+{{n}^{3}}\]

Cancel out similar terms \[m{{n}^{2}}\]and \[{{m}^{2}}n\].

\[={{m}^{3}}+{{n}^{3}}\]

\[\therefore \left( m+n \right)\left( {{m}^{2}}-mn+{{n}^{2}} \right)={{m}^{3}}+{{n}^{3}}\]

Hence, we have found the product of the algebraic expression.

Note:

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\[{{m}^{3}}+{{n}^{3}}\].

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,

\[\begin{align}

& \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] \\

\end{align}\]

Opening the brackets and simplifying it,

\[={{m}^{3}}-{{m}^{2}}n+m{{n}^{2}}+n{{m}^{2}}-m{{n}^{2}}+{{n}^{3}}\]

Cancel out similar terms \[m{{n}^{2}}\]and \[{{m}^{2}}n\].

\[={{m}^{3}}+{{n}^{3}}\]

\[\therefore \left( m+n \right)\left( {{m}^{2}}-mn+{{n}^{2}} \right)={{m}^{3}}+{{n}^{3}}\]

Hence, we have found the product of the algebraic expression.

Note:

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\[{{m}^{3}}+{{n}^{3}}\].

Last updated date: 29th Sep 2023

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