# Expansion of brackets

• April 18th 2008, 05:27 PM
AS1
Expansion of brackets

• April 18th 2008, 05:58 PM
Mathstud28
Quote:

Originally Posted by AS1

the answer is $
x^3+x^2y+\frac{1}{3}xy^2+\frac{1}{27}y^3$

just expand it

or use Pascal's Triangle

Or use the Binomial theorem

Mathstud..
• April 18th 2008, 06:01 PM
Soroban
Hello, AS1!

Have you never cubed anything before?

Quote:

Expand and simplify: .(3x-y/3)^3
I'm not sure exactly what the problem says.
. . I'll do it both ways . . .

$\left(\frac{3x-y}{3}\right)^3 \;=\;\frac{(3x-y)^3}{3^3} \;= \;\frac{(3x)^3 - 3(3x)^2y + 3(3x)y^2 - y^3}{27}$

. . . . . . . $= \;\frac{27x^3 - 27x^2y + 9xy^2 - y^3}{27}$

$\left(3x-\frac{y}{3}\right)^3 \;=\;(3x)^3 - 3(3x)^2\left(\frac{y}{3}\right) + 3(3x)\left(\frac{y}{3}\right)^2 - \frac{y^3}{27}$

. . . . . . . $= \;27x^3 - 9x^2y + xy^2 - \frac{y^3}{27}$