Question

Find the square of $2a-b-3c$ .
(a) $4{{a}^{2}}+{{b}^{2}}+9{{c}^{2}}-4ab+6bc-ac$
(b) $4{{a}^{2}}+{{b}^{2}}+9{{c}^{2}}-4ab+bc-12ac$
(c) $4{{a}^{2}}+{{b}^{2}}+9{{c}^{2}}-4ab+6bc-12ac$
(d) $4{{a}^{2}}+{{b}^{2}}+9{{c}^{2}}-4ab+6bc-12ac$

Answer 1

Hint: In the given expression consider (2a - b) as one term and 3c as second term. Apply the algebraic identity ${{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$ to expand the square of $2a-b-3c$. Now, simplify the expression obtained and again apply the same algebraic identity to expand the square of (2a – b). Check the options matching the obtained expression to get the answer.

Complete step-by-step solution:
Here, we have been provided the expression $2a-b-3c$ and we are asked to find the square of this expression, i.e. ${{\left( 2a-b-3c \right)}^{2}}$.
Now let us assume the value of the required expression as E. So, we have
$\Rightarrow E={{\left( 2a-b-3c \right)}^{2}}$
Here, we can see that we have to find the square of three terms, 2a, b, and 3c, each separated by a minus (-) sign. We know that the formula for square of difference of two terms is given by the algebraic identity ${{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$. So, let us convert the three terms given in expression E into two terms so that we can apply the above mentioned algebraic identity.
Now, grouping (2a – b) and considering it as a single term, we get the required expression as
$\Rightarrow E={{\left[ \left( 2a-b \right)-3c \right]}^{2}}$
Now applying the algebraic identity ${{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$ by considering 2a – b = x and 3c = y, we get
$\begin{align}
  & \Rightarrow E={{\left( 2a-b \right)}^{2}}-2\times \left( 2a-b \right)\times 3c+{{\left( 3c \right)}^{2}} \\
 & \Rightarrow E={{\left( 2a-b \right)}^{2}}-6c\left( 2a-b \right)+9{{c}^{2}} \\
 & \Rightarrow E={{\left( 2a-b \right)}^{2}}-12ac+6bc+9{{c}^{2}} \\
\end{align}$
Applying the same algebraic identity for ${{\left( 2a-b \right)}^{2}}$ , we get
$\begin{align}
  & \Rightarrow E={{\left( 2a \right)}^{2}}-2\times 2a\times b+{{b}^{2}}-12ac+6bc+9{{c}^{2}} \\
 & \Rightarrow E=4{{a}^{2}}-4ab+{{b}^{2}}-12ac+6bc+9{{c}^{2}} \\
 & \Rightarrow E=4{{a}^{2}}+{{b}^{2}}+9{{c}^{2}}-4ab+6bc-12ac \\
\end{align}$
Hence, option (c) is the correct answer.

Note: One may note that we have a direct formula for the expansion of square of three terms given as ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz$. So, you can remember the direct formula to get the answer. Here, you may note that we have not applied the direct formula but we have used a general method. We can memorize the square of three terms only because books do not provide the expansion formula for squares of 4 or more terms. In that case, you have to apply the general method that we have used above. Now, in the above solution, we have grouped (2a – b), you may group any two terms as the answer will remain the same.