Question

Find $ a $ if the coefficients of $ {x^2} $ and $ {x^3} $ in the expansion of $ {\left( {3 + ax} \right)^3} $ are equal.

Answer 1

Hint: In this problem, first we will write expansion of $ {\left( {3 + ax} \right)^3} $ by using the formula $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $ . Then, we will equate the coefficients of $ {x^2} $ and $ {x^3} $ . We will get the required value of $ a $ .

Complete step-by-step answer:
In this problem, we need to find the value of $ a $ . It is given that coefficients of $ {x^2} $ and $ {x^3} $ are equal in the expansion of $ {\left( {3 + ax} \right)^3} $ . So, first we will write the expansion of $ {\left( {3 + ax} \right)^3} $ .
We know that expansion of $ {\left( {a + b} \right)^3} $ is given by $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $ . Use this information and let us write the expansion of $ {\left( {3 + ax} \right)^3} $ . Therefore, we get
 $ {\left( {3 + ax} \right)^3} = {\left( 3 \right)^3} + {\left( {ax} \right)^3} + 3\left( 3 \right)\left( {ax} \right)\left( {3 + ax} \right) \cdots \cdots \left( 1 \right) $
Let us simplify the RHS of the equation $ \left( 1 \right) $ . So, we get
 $
  {\left( {3 + ax} \right)^3} = 27 + {a^3}{x^3} + 9ax\left( {3 + ax} \right) \\
   \Rightarrow {\left( {3 + ax} \right)^3} = 27 + {a^3}{x^3} + 27ax + 9{a^2}{x^2} \cdots \cdots \left( 2 \right) \\
  $
From equation $ \left( 2 \right) $ , we can say that the coefficient of $ {x^3} $ is $ {a^3} $ and the coefficient of $ {x^2} $ is $ 9{a^2} $ . In the problem, it is given that coefficients of $ {x^2} $ and $ {x^3} $ are equal. So, let us equate the coefficients of $ {x^2} $ and $ {x^3} $ . Therefore, we get $ 9{a^2} = {a^3} \cdots \cdots \left( 3 \right) $ . Now we are going to solve the equation $ \left( 3 \right) $ to find the required value of $ a $ . Therefore, from equation $ \left( 3 \right) $ , we get
 $
  9{a^2} - {a^3} = 0 \\
   \Rightarrow {a^2}\left( {9 - a} \right) = 0 \\
  $
 $ \Rightarrow {a^2} = 0 $ or $ 9 - a = 0\quad \left[ {\because ab = 0 \Rightarrow a = 0} \right. $ or $ \left. {b = 0} \right] $
  $ \Rightarrow a = 0 $ or $ a = 9 $
Therefore, we can say that if the coefficients of $ {x^2} $ and $ {x^3} $ in the expansion of $ {\left( {3 + ax} \right)^3} $ are equal then there are two possible values of $ a $ . That is, $ a = 0 $ or $ a = 9 $ .

Note: In this problem, we expand $ {\left( {3 + ax} \right)^3} $ by using the formula $ {\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) $ . Note that here the power of $ \left( {3 + ax} \right) $ is $ 3 $ . Assume that the power of $ \left( {3 + ax} \right) $ is higher than $ 3 $ then we will use the expansion of $ {\left( {a + b} \right)^n} $ . For example, if we want to expand $ {\left( {3 + ax} \right)^9} $ then we will use the expansion of $ {\left( {a + b} \right)^n} $ which is given by $ {\left( {a + b} \right)^n} = {}^n{C_0}{\left( a \right)^n}{b^0} + {}^n{C_1}{\left( a \right)^{n - 1}}{b^1} + {}^n{C_2}{\left( a \right)^{n - 2}}{b^2}... + {}^n{C_n}{\left( a \right)^{n - n}}{b^n} $