exponents

Question

Can someone please explain how this simplified formula is derived:

(x+y)^2=x^2+2xy+y^2

Also, would this be correct....

(x+y)^3=x^3+3xy+y^3

IanStewart · Answer

One way to expand an expression like (x+y)^2 is to do as follows:
 \begin{align}
(x+y)^2 &= (x+y)(x+y) \
&= x(x+y) + y(x+y) \
&= x^2 + xy + xy + y^2 \
&= x^2 + 2xy + y^2
\end{align}

So the second expansion you've suggested is not correct. Instead,

\begin{align}
(x+y)^3 &= (x+y) (x+y)^2 \ 
&= (x+y)(x^2 + 2xy + y^2) \
&= x(x^2 + 2xy + y^2) + y(x^2 + 2xy + y^2) \
&= x^3 + 2x^2y + xy^2 + x^2y + 2xy^2 + y^3 \
&= x^3 + 3x^2y + 3xy^2 + y^3
\end{align}

You don't often need to expand a cube on the GMAT, fortunately, but the (x+y)^2 pattern shows up very often.

golantrevize · Answer

napy27, a few more formulae for your reference -

(a-b)^2 = a^2 - 2ab-b^2
(a+b)(a-b) = a^2 - b^2 
(a-b)^3 = a^3 -3(a^2)b+3a(b^2)-b^3
All of these formulae can be derived by multiplying the elements of the parentheses, one by one.
I suggest you learn these formulae by rote. In case you can't remember them while taking a test, try using substitutions for "a" and "b" to get to the answer.

golantrevize · Answer

The generic formula is this -

(a + b)^n = nC0 a^n + nC1 a^(n − 1)b + nC2 a^(n − 2)b² + nC3 a^(n − 3)b^3 + ......b^n

Google binomial expansion or check this page out - https://www.themathpage.com/aprecalc/bin ... heorem.htm
You can also look for Pascal triangle for an easy and quick way to get the expansions.

amit2k9 · Answer

use pascal triangle method ---

1
121
1331
14641 and so on.

so, x^2+ 2x+1

x^3 + 3x^2+ 3^x +1

x^4 + 4x^3+ 6x^2+4x+ 1

notice that powers of x decreases from left to right.

exponents

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