Question

Using identities, evaluate ${{998}^{2}}$.
(a) 995704
(b) 996004
(c) 992004
(d) 990004

Answer 1

Hint: Assume the value of the given expression as E. Now, write the base number 998 as the difference of two numbers 1000 and 2. Use the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and substitute the value a = 1000 and b = 2 to expand the terms. Perform simple addition and subtraction to get the answer.

Complete step by step answer:
Here we have been provided with the expression ${{998}^{2}}$ and we are asked to simplify it using the suitable algebraic identities. Assuming the given expression as E we have,
$\Rightarrow E={{998}^{2}}$
Now, we can write the base number which is 998 as the difference of two numbers which are 1000 and 2, so we get,
$\Rightarrow E={{\left( 1000-2 \right)}^{2}}$
We know that the whole square formula of the difference of two numbers is given as ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, so considering a = 1000 and b = 2 we get,
$\begin{align}
  & \Rightarrow E={{\left( 1000 \right)}^{2}}+{{2}^{2}}-2\left( 1000 \right)\left( 2 \right) \\
 & \Rightarrow E=1000000+4-4000 \\
\end{align}$
Performing simple addition and subtraction we get,
$\therefore E=996004$

So, the correct answer is “Option b”.

Note: Do not write the given base 998 as the sum of two numbers like 996 + 2 etc. because then it would be difficult to calculate the square of 996 just like it is difficult to calculate the square of 998 directly as it is a large number. You must remember the other two important algebraic identities given as ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$. In addition to these we also have certain algebraic identities involving the cube of numbers that must be remembered like ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ and ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$.