Chapter 5.5, Problem 65WE

To determine

To prove: that if square of the difference of two numbers is subtracted from the square of their sum, then it is equal to four times their product.

Explanation of Solution

Formula Used:

  $a^2-b^2=\left(a+b\right)\left(a-b\right)$

Calculation:

Let the two numbers be x and y .

Then, the square of their sum is: ${\left(x+y\right)}^2$

Then, the square of their difference is: ${\left(x+y\right)}^2$

Now, the square of the sum of these two numbers minus the square of their difference is:

  $\begin{array}{c}{\left(x+y\right)}^2-{\left(x-y\right)}^2=\left(\left(x+y\right)+\left(x-y\right)\right)\left(\left(x+y\right)-\left(x-y\right)\right)\\ =\left(x+y+x-y\right)\left(x+y-x+y\right)\\ =\left(2x\right)\left(2y\right)\\ =4xy\end{array}$

Hence proved