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# Find the product of$\left( m+n \right)\left( {{m}^{2}}-mn+{{n}^{2}} \right)$. Verified
362.4k+ views
Hint:Given is an algebraic expression. Open the brackets and simplify the expression. You will get the product of the expression.

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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