
What is the formula for \[{{a}^{3}}-{{b}^{3}}\]?
Answer
553.5k+ views
Hint: Consider the difference of two numbers a and b given as (a – b). Now, take the product of this expression, with itself three times. Write the L.H.S in the exponent form as \[{{\left( a-b \right)}^{3}}\] and multiply each term of \[\left( a-b \right)\times \left( a-b \right)\times \left( a-b \right)\], simplify them. Leave \[{{a}^{3}}-{{b}^{3}}\] at one side and take all other terms to the other side to get the answer.
Complete step by step answer:
Here, we have been asked to derive a formula for \[{{a}^{3}}-{{b}^{3}}\].
Now, we can clearly see that the power of both the terms ‘a’ and ‘b’ in the expression \[{{a}^{3}}-{{b}^{3}}\] is 3 and there is a minus sign between the terms. So, let us consider the difference (a – b). Now, multiplying (a – b) three times with itself, we have,
\[\begin{align}
& \Rightarrow \left( a-b \right)\left( a-b \right)\left( a-b \right)={{\left( a-b \right)}^{3}} \\
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( a-b \right)\left( a-b \right)\left( a-b \right) \\
\end{align}\]
Multiplying the terms in the R.H.S by taking two at a time, we get,
\[\begin{align}
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{2}}-ab-ba+{{b}^{2}} \right)\left( a-b \right) \\
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{2}}-2ab+{{b}^{2}} \right)\left( a-b \right) \\
\end{align}\]
Now, multiplying and simplifying the remaining term in the R.H.S, we get,
\[\begin{align}
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-{{a}^{2}}b-2{{a}^{2}}b+2{{a}^{2}}b+{{b}^{2}}a-{{b}^{3}} \right)\left( a-b \right) \\
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} \right) \\
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-{{b}^{3}} \right)+3a{{b}^{2}}-3{{a}^{2}}b \\
& \Rightarrow {{\left( a-b \right)}^{3}}={{\left( a-b \right)}^{3}}-3a{{b}^{2}}+3{{a}^{2}}b \\
& \Rightarrow {{\left( a-b \right)}^{3}}={{\left( a-b \right)}^{3}}+3ab\left( a-b \right) \\
\end{align}\]
Therefore, the above expression is the required formula for the provided expression \[\left( {{a}^{3}}-{{b}^{3}} \right)\].
Note:
One may note that the expression we have obtained for \[\left( {{a}^{3}}-{{b}^{3}} \right)\] in the last part of the solution, can be written in different forms also. If we will take (a – b) common and then simplify the R.H.S. in the end part then we will get another formula given as: - \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]. Now, you may check the formula if it is correct or not by substituting some numerical values of ‘a’ and ‘b’. If both the L.H.S and the R.H.S turn out to be equal then our derived formula is correct otherwise not. You may also use the method of the binomial theorem to solve the question.
Complete step by step answer:
Here, we have been asked to derive a formula for \[{{a}^{3}}-{{b}^{3}}\].
Now, we can clearly see that the power of both the terms ‘a’ and ‘b’ in the expression \[{{a}^{3}}-{{b}^{3}}\] is 3 and there is a minus sign between the terms. So, let us consider the difference (a – b). Now, multiplying (a – b) three times with itself, we have,
\[\begin{align}
& \Rightarrow \left( a-b \right)\left( a-b \right)\left( a-b \right)={{\left( a-b \right)}^{3}} \\
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( a-b \right)\left( a-b \right)\left( a-b \right) \\
\end{align}\]
Multiplying the terms in the R.H.S by taking two at a time, we get,
\[\begin{align}
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{2}}-ab-ba+{{b}^{2}} \right)\left( a-b \right) \\
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{2}}-2ab+{{b}^{2}} \right)\left( a-b \right) \\
\end{align}\]
Now, multiplying and simplifying the remaining term in the R.H.S, we get,
\[\begin{align}
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-{{a}^{2}}b-2{{a}^{2}}b+2{{a}^{2}}b+{{b}^{2}}a-{{b}^{3}} \right)\left( a-b \right) \\
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} \right) \\
& \Rightarrow {{\left( a-b \right)}^{3}}=\left( {{a}^{3}}-{{b}^{3}} \right)+3a{{b}^{2}}-3{{a}^{2}}b \\
& \Rightarrow {{\left( a-b \right)}^{3}}={{\left( a-b \right)}^{3}}-3a{{b}^{2}}+3{{a}^{2}}b \\
& \Rightarrow {{\left( a-b \right)}^{3}}={{\left( a-b \right)}^{3}}+3ab\left( a-b \right) \\
\end{align}\]
Therefore, the above expression is the required formula for the provided expression \[\left( {{a}^{3}}-{{b}^{3}} \right)\].
Note:
One may note that the expression we have obtained for \[\left( {{a}^{3}}-{{b}^{3}} \right)\] in the last part of the solution, can be written in different forms also. If we will take (a – b) common and then simplify the R.H.S. in the end part then we will get another formula given as: - \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]. Now, you may check the formula if it is correct or not by substituting some numerical values of ‘a’ and ‘b’. If both the L.H.S and the R.H.S turn out to be equal then our derived formula is correct otherwise not. You may also use the method of the binomial theorem to solve the question.
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