Question

Which of the following identity can be used to solve 1001 $\times$ 999?

• A. $(a+b)\times (a-b)=a^2$ - $b^2$
• B. ${(a+b)}^2$ = $a^2$ + $2ab$ + $b^2$
• C. ${(a-b)}^2$ = $a^2$ -$2ab$ + $b^2$
• D. $(x+y{)}^3$ = $x^3$ + 3$x^2$$y$ + 3$x$$y^2$ + $y^3$

Answer 1

The correct option is A $(a+b)\times (a-b)=a^2$ - $b^2$

1001 = (1000 + 1)
999 = (1000 - 1)
(1000 + 1)(1000 - 1) = ${1000}^2$ - $1^2$ [Identity (a + b) $\times$ (a - b) = $a^2$ - $b^2$ is used]
= 1000000 -1 = 999999