Question

If a – b = 3 and ab = 10, then using algebraic identity, find the value of ${{a}^{3}}-{{b}^{3}}=$?

Answer 1

Hint: Use the algebraic identity of ${{\left( a-b \right)}^{3}}$, which given as ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$. Get the value of ${{a}^{3}}-{{b}^{3}}$from above expression and use the given values of (a - b) and ab to simplify it further.

Complete step-by-step answer:
We are given equations in a and b as
a – b = 3…………… (i)
ab = 10…………….(ii)
And using the above equations, we need to find the value of ${{a}^{3}}-{{b}^{3}}=?$
As we know the algebraic identity of ${{\left( a-b \right)}^{3}}$ can be given as
${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$………….. (iii)
Now, we can observe the identity (iii) and get that the right hand side of the identity is consists of expression ${{a}^{3}}-{{b}^{3}}$ and hence, we can get value of it by putting the values of (a – b) and ab from the equation (i) and (ii). So, we can express the equation (iii) by transferring the term -3ab (a – b) to other side for getting value of ${{a}^{3}}-{{b}^{3}}$as
$\begin{align}
  & {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}=3ab\left( a-b \right), \\
 & {{\left( a-b \right)}^{3}}+3ab\left( a-b \right)={{a}^{3}}-{{b}^{3}}, \\
\end{align}$
${{a}^{3}}-{{b}^{3}}={{\left( a-b \right)}^{3}}+3ab\left( a-b \right)$………….. (iv)
Hence, now we get the value of ${{a}^{3}}-{{b}^{3}}$ in terms of (a – b) and ab only. And as we know he values of
(a – b) and ab as 3 and 10 respectively. So, we can put these values to the expression (iv) to get the value of ${{a}^{3}}-{{b}^{3}}$. Hence, on putting values of (a – b) and ab to the equation (iv), we get
${{a}^{3}}-{{b}^{3}}={{\left( 3 \right)}^{3}}+3\times 10\left( 3 \right)$
Now, we know ${{\left( 3 \right)}^{3}}$ can be written as products of the number ‘3’ to three times. So, the value of $\left( {{3}^{3}}=3\times 3\times 3=27 \right)$ would be 27. Hence, we can get value of ${{a}^{3}}-{{b}^{3}}$as
 ${{a}^{3}}-{{b}^{3}}=27+90=117$
Hence, the value of ${{a}^{3}}-{{b}^{3}}$is 117, if the value of a – b and ab are 3 and 10 respectively.

Note: One may use another identity of ${{a}^{3}}-{{b}^{3}}$in the following approach as well.
 So, another identity of ${{a}^{3}}-{{b}^{3}}$ is given as
 ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
Now, use identity of ${{\left( a-b \right)}^{2}}$, which is given as:
$\begin{align}
  & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab, \\
 & {{a}^{2}}+{{b}^{2}}={{\left( a-b \right)}^{2}}+2ab \\
\end{align}$
 Now, replace ${{a}^{2}}+{{b}^{2}}$by the above expression in the identity of ${{a}^{3}}-{{b}^{3}}$. So, we get
${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{\left( a-b \right)}^{2}}+2ab+ab \right)$
Now, put values of $\left( a-b \right)$ and ab to the above expression.
One may find the exact values of a and b as well by solving the expression a – b = 3 and ab = 10. So, replace, a by b + 3 to the expression ab = 10. So, we get
b(b + 3) = 10
$\begin{align}
  & {{b}^{2}}+3b-10=0 \\
 & {{b}^{2}}+5b-2b-10=0 \\
\end{align}$
b(b + 5) – 2 (b + 5) = 0
b = 2 and b = -5
And find corresponding values of ‘a’ as well. So, it can be another approach for the problem as well.