Let \( a = b \).

Then,
\[ \begin{array} {c r c l }

\color{red} {1} & ab & = & a^2 \\
\color{orange} {2} & ab - b^2 & = & a^2 -b^2 \\
\color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\
\color{green} {4} & b & = & a+b \\
\color{blue} {5} & b & = & b+b \\
\color{indigo} {6} & b & = & 2b \\
\color{purple} {7} & 1 & = & 2.
\end{array} \]
Which line is the **first** that contains an error in this proof?

\color{red} {1} & ab & = & a^2 \\
\color{orange} {2} & ab - b^2 & = & a^2 -b^2.
\end{array} \]
\(b^2\) is subtracted from both sides of the equal sign. Is it equally valid to do the operation below instead?
\[ \begin{array} {c r c l }

\color{red} {1} & ab & = & a^2 \\
\color{orange} {2} & b^2 - ab & = & b^2 - a^2 \\
\end{array} \]

\[ \begin{array} {c r c l }

\color{orange} {2} & ab - b^2 & = & a^2 -b^2 \\
\color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\
\end{array} \]

On the left-hand side of line 3, the value \(b\) is "factored out" from \( ab-b^2 \) to get \( b(a-b) .\)

Is the equation below true? \[ ab - b^2 - a^2 = ab(1-b-a) \]

\[ \begin{array} {c r c l }

\color{orange} {2} & ab - b^2 & = & a^2 -b^2 \\
\color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\
\end{array} \]

Changing \( a^2 - b^2 \) to \( (a+b)(a-b) \) is known as the difference of squares identity (it was first introduced in a previous course, as shown in the video below). What if we started with fourth powers instead, that is, \( a^4 - b^4 ?\) Which expression is equivalent?

\color{gold} {3} & b(a-b) & = & (a+b)(a-b) \\
\color{green} {4} & b & = & a+b.
\end{array} \]
\( (a-b) \) does what is known as a "cancel" on both sides of the equal sign. Since the problem set up \( a = b,\) this implies division by zero. Suppose that \( a \neq b \) instead; what identity would be used to cancel? \((\)Assume \(Q \neq 0.)\)

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