Question 818532
Short answer: No<br>
{{{((d+e)-9)^2  }}}
There are a couple of ways to figure this out:<ul><li>Multiply it out:<ol><li>Rewrite as {{{((d+e)-9)((d+e)-9)}}}</li><li>Multiply each term of the first (d+e)-9 times each term of the second (d+e)-9. This will be 9 multiplications!</li><li>Add like terms, if any.</li></ul></li><li> Use patterns:<ul><li>{{{(a+b)^2= a^2+2ab+b^2}}}</li><li>{{{(a-b)^2= a^2-2ab+b^2}}}</li></ul></li></ul>I am going to show you how to use the patterns.<br>
At first you might think: The patterns are for squaring two-term expressions but I'm trying to square a three-term expression. How is that possible?<br>
Well, the problem already has the d+e grouped. We can treat it as the "a" in {{{(a-b)^2}}}. Using (d+e) as the "a" and the "9" as the "b", the pattern shows us how to square your expression:
{{{(d+e)^2-2(d+e)(9)+(9)^2  }}}
If you can't see the pattern in this, then check out the P.S. at the end. Simplifying the last parts:
{{{(d+e)^2-18(d+e)+81 }}}
{{{(d+e)^2-18d-18e+81 }}}
To simplify the beginning we can use the {{{(a+b)^2}}} pattern (in a more obvious way than above):
{{{(d)^2+2(d)(e)+(e)^2-18d-18e+81 }}}
which simplifies:
{{{d^2+2de+e^2-18d-18e+81 }}}
There are no like terms. So the expression is fully simplified.<br>
P.S. If you're having trouble with how we used the {{{(a-b)^2}}} pattern on {{{((d+e)-9)^2  }}} then using a temporary variable can help:
Let q = (d+e). Substituting this in for (d+e) we get:
{{{(q-9)^2  }}}
Use of the {{{(a-b)^2}}} pattern should be clear now:
{{{(q)^2-2(q)(9)+(9)^2}}}
which simplifies:
{{{q^2-18q+81}}}
Now we need to replace the "q" with "(d+e)":
{{{(d+e)^2-18(d+e)+81}}}
which is the same as we had earlier. It simplifies to the same answer we got above: {{{d^2+2de+e^2-18d-18e+81 }}}<br>
P.P.S. If you insist on multiplying {{{((d+e)-9)^2}}} out instead of using the patterns, you should end up with the same result.