Question

Prove that:
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$

Answer 1

Hint: To prove the identity given in the above question, we have to firstly consider the LHS, which is equal to ${{\left( a+b \right)}^{2}}$. Since the power over an expression indicates the number of times it is being multiplied. Therefore, the LHS will be equal to $\left( a+b \right)\left( a+b \right)$. For simplifying it, we need to apply the distributive law of the algebraic multiplication which is given by \[a\left( b+c \right)=ab+ac\]. On applying it repeatedly, we will get the LHS equal to RHS and hence the given identity will get proved.

Complete step-by-step answer:
The identity to be proved given in the above question is
$\Rightarrow {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}.......\left( i \right)$
Let us consider the LHS of the above identity.
$\Rightarrow LHS={{\left( a+b \right)}^{2}}$
We know that the square over a number is equal to the number multiplied by itself. Therefore, we can write the above expression as
$\Rightarrow LHS=\left( a+b \right)\left( a+b \right)$
Now, applying the distributive law of the algebraic multiplication, which is givne by \[a\left( b+c \right)=ab+ac\], we can write the above expression as
$\Rightarrow LHS=\left( a+b \right)a+\left( a+b \right)b$
Now, by using the commutative property of multiplication, we can write the above expression as
$\Rightarrow LHS=a\left( a+b \right)+b\left( a+b \right)$
Now, again using the distributive law of multiplication, we can write the above expression as
$\begin{align}
  & \Rightarrow LHS=a\left( a \right)+ab+ba+b\left( b \right) \\
 & \Rightarrow LHS={{a}^{2}}+ab+ab+{{b}^{2}} \\
 & \Rightarrow LHS={{a}^{2}}+2ab+{{b}^{2}}........\left( ii \right) \\
\end{align}$
From (i) we can write
$\Rightarrow RHS={{a}^{2}}+2ab+{{b}^{2}}.......\left( iii \right)$
Finally, on from the equations (ii) and (iii) we can write
$\Rightarrow LHS=RHS$
Hence, we have proved the given identity.

Note: We can prove the given identity by considering the RHS also. The RHS is equal to ${{a}^{2}}+2ab+{{b}^{2}}$ which can be modified by splitting the middle term as ${{a}^{2}}+ab+ab+{{b}^{2}}$. Then taking a common from the first two terms and b common from the last two terms, we will get \[a\left( a+b \right)+b\left( a+b \right)\]. Finally, taking \[\left( a+b \right)\] common, we will get $\left( a+b \right)\left( a+b \right)$ which can be written as ${{\left( a+b \right)}^{2}}$, which is the LHS.