What is the formula for ${{\left( a-b-c \right)}^{2}}$ ?

Hint: To find the formula for ${{\left( a-b-c \right)}^{2}}$ , we have to write the given expression without exponents. Then, we have to apply distributive property and FOIL methods. We have to apply the commutative property. Then, we have to combine the like terms and add.

Complete step by step answer:
We have to find the formula for ${{\left( a-b-c \right)}^{2}}$ . Let us write the given expression without exponents.
$\Rightarrow {{\left( a-b-c \right)}^{2}}=\left( a-b-c \right)\left( a-b-c \right)$
Now, we have to apply the distributive property and FOIL method. We can write the above equation as
\[\Rightarrow {{\left( a-b-c \right)}^{2}}=\left( a\times a \right)+\left( a\times -b \right)+\left( a\times -c \right)+\left( -b\times a \right)+\left( -b\times b \right)+\left( -b\times -c \right)+\left( -c\times a \right)+\left( -c\times -b \right)+\left( -c\times -c \right)\]
Now, let us multiply the terms inside the brackets.
$\Rightarrow {{\left( a-b-c \right)}^{2}}={{a}^{2}}-ab-ac-ba+{{b}^{2}}+bc-ca+cb+{{c}^{2}}$
We have to apply the commutative property.
$\Rightarrow {{\left( a-b-c \right)}^{2}}={{a}^{2}}-ab-ac-ab+{{b}^{2}}+bc-ac+bc+{{c}^{2}}$
Now, let us combine the like terms.
$\Rightarrow {{\left( a-b-c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-ab+bc+bc-ac-ac$
We have to add the like terms. We can write the result of this operation as
$\Rightarrow {{\left( a-b-c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2ab+2bc-2ac$
Therefore, the formula for ${{\left( a-b-c \right)}^{2}}$ is ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2ab+2bc-2ac$ .

Note: Students must be thorough with algebraic properties. Here, we have used distributive and commutative property. Distributive property states $a\left( b+c \right)=ab+bc$ . Commutative property states that $ab=ba$ . Let us see what the FOIL method is. The FOIL method is a special case of a more general method for multiplying algebraic expressions using the distributive law. We can write the formula for the FOIl method as $\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd$ . We can also find the formula of ${{\left( a-b-c \right)}^{2}}$ using the formula of ${{\left( a+b+c \right)}^{2}}$ .
We know that ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz...\left( i \right)$ .
Let us write ${{\left( a-b-c \right)}^{2}}$ as
${{\left( a-b-c \right)}^{2}}={{\left[ a+\left( -b \right)+\left( -c \right) \right]}^{2}}$
When we compare the above form to (i), we will get $x=a,y=-b,z=-c$ . Let us substitute these values in equation (i).
$\begin{align}
  & \Rightarrow {{\left[ a+\left( -b \right)+\left( -c \right) \right]}^{2}}={{a}^{2}}+{{\left( -b \right)}^{2}}+{{\left( -c \right)}^{2}}+2a\left( -b \right)+2\left( -b \right)\left( -c \right)+2a\left( -c \right) \\
 & \Rightarrow {{\left( a-b-c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2ab+2bc-2ac \\
\end{align}$