Question

Find the expansion of ${{\left( 2a-3b \right)}^{3}}$ :
(a) $8{{a}^{3}}-27{{b}^{3}}-36{{a}^{2}}b-54a{{b}^{2}}$
(b) $8{{a}^{3}}+27{{b}^{3}}-36{{a}^{2}}b+54a{{b}^{2}}$
(c) $8{{a}^{3}}-27{{b}^{3}}+36{{a}^{2}}b+54a{{b}^{2}}$
(d) $8{{a}^{3}}-27{{b}^{3}}-36{{a}^{2}}b+54a{{b}^{2}}$

Answer 1

Hint: Apply the binomial expansion of ${{\left( a-b \right)}^{3}}$ that you get by using the general expansion of ${{\left( a-b \right)}^{n}}$. Finally, replace ‘a‘ by 2a and ‘b’ by 3b to get the expansion of ${{\left( 2a-3b \right)}^{3}}$.

Complete step-by-step solution -
We know that the binomial expansion of ${{\left( a-b \right)}^{n}}$ , can be written as:
${{\left( a-b \right)}^{n}}={{\text{ }}^{n}}{{\text{C}}_{0}}{{a}^{n}}{{b}^{0}}{{-}^{n}}{{\text{C}}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{\text{C}}_{2}}{{a}^{n-2}}{{b}^{2}}-.........{{+}^{n}}{{\text{C}}_{n}}{{a}^{0}}{{\left( -b \right)}^{n}}$
Therefore, if we take n to be 3, the binomial expansion of ${{\left( a-b \right)}^{3}}$ is:
${{\left( a-b \right)}^{3}}={{\text{ }}^{3}}{{\text{C}}_{0}}{{a}^{3}}{{b}^{0}}{{-}^{3}}{{\text{C}}_{1}}{{a}^{2}}{{b}^{1}}{{+}^{3}}{{\text{C}}_{2}}{{a}^{1}}{{b}^{2}}{{-}^{3}}{{\text{C}}_{3}}{{a}^{0}}{{b}^{3}}$
Now, if we replace ‘a’ by 2a and ‘b’ by 3b, in the above expansion, we will end up getting the expansion of ${{\left( 2a-3b \right)}^{3}}$ .
$\therefore {{\left( 2a-3b \right)}^{3}}={{\text{ }}^{3}}{{\text{C}}_{0}}{{(2a)}^{3}}{{(3b)}^{0}}{{-}^{3}}{{\text{C}}_{1}}{{(2a)}^{2}}{{(3b)}^{1}}{{+}^{3}}{{\text{C}}_{2}}{{(2a)}^{1}}{{(3b)}^{2}}{{-}^{3}}{{\text{C}}_{3}}{{(2a)}^{0}}{{(3b)}^{3}}$
$\Rightarrow {{\left( a-b \right)}^{3}}=\text{ 8}{{\times }^{3}}{{\text{C}}_{0}}{{a}^{3}}{{b}^{0}}-12{{\times }^{3}}{{\text{C}}_{1}}{{a}^{2}}{{b}^{1}}+18{{\times }^{3}}{{\text{C}}_{2}}{{a}^{1}}{{b}^{2}}-27{{\times }^{3}}{{\text{C}}_{3}}{{a}^{0}}{{b}^{3}}$
Now we know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . Therefore, our equation becomes:
${{\left( a-b \right)}^{3}}=\text{ 8}\times \dfrac{3!}{3!0!}{{a}^{3}}{{b}^{0}}-12\times \dfrac{3!}{2!1!}{{a}^{2}}{{b}^{1}}+18\times \dfrac{3!}{2!1!}{{a}^{1}}{{b}^{2}}-27\times \dfrac{3!}{3!0!}{{a}^{0}}{{b}^{3}}$
We also know that 0! Is equal to 1.
 $\therefore {{\left( a-b \right)}^{3}}=\text{ 8}{{a}^{3}}-36{{a}^{2}}{{b}^{1}}+54{{a}^{1}}{{b}^{2}}-27{{b}^{3}}$
Therefore, we can conclude that the answer to the above question is option (d).

Note: Always be careful with the signs that appear in the expansions, as the students are generally finding signs to be a concern while using the binomial expansions. Also, be careful about the calculation part, as in general cases, the questions involving binomial expansion contain very long and complex calculations due to the presence of factorial terms. You should also know that the binomial coefficient and actual coefficients might or might not be the same. For example: in the expansion of ${{\left( 1+3x \right)}^{3}}$ , the binomial coefficient of ${{x}^{3}}$ is $^{3}{{C}_{3}}=1$ and coefficient is 27.