How would one solve (x-y)^6?

Question

So far, I've figured out: (x-y)(x-y)(x-y)(x-y)(x-y)(x-y), but I'm not sure where to go from here.

Excuse my grammar. I meant "simplify," not solve. No need to be rude. :)

*cough* Wrong again! Haha. :)
How would one multiply (x-y)^6?

Answer 1

You can of course, simply expand the terms in parentheses one at a time using the distributive property, like so:

x(x-y)^5 - y(x-y)^5
x²(x-y)^4 - xy(x-y)^4 - xy(x-y)^4 + y²(x-y)^4 ...

And so on. This is fairly tedious however. Fortunately, there is an easier way. If you examine what happens to the coefficients of (x+y)^n as n increases you find a nice pattern:

x+y -- coefficients 1, 1
(x+y)² = x² + 2xy + y² -- coefficients 1 2 1
(x+y)³ = (x+y)(x²+2xy+y²) = x³ + 3x²y + 3xy² + y³ -- coefficients 1 3 3 1
(x+y)^4 = (x+y)(x³+3x²y+3xy²+y³) = x^4 + 4x³y + 6x²y² + 4xy³ + 1 -- coefficients 1 4 6 4 1

Arranging these coefficients in a triangle, we obtain:

..... 1 1
.....1 2 1
....1 3 3 1
...1 4 6 4 1

As you can see, each number in the triangle is the sum of the two numbers immediately above it. The triangle generated by this pattern is known as Pascal's triangle (yes, they named a triangle after me. Or at least, after the guy whose name I'm using.) It can easily be shown that the pattern of coefficients continues indefinitely. Therefore, all we need do to find (x+y)^6 is generate Pascal's triangle out to the sixth row, and place those coefficients before the corresponding powers of x and y. So continuing from the fourth row:

...1 .4 ...6 ..4 1
..1 5 .10 10. 5 1
1 6 15 20 15 6 1

So placing these coefficients before the appropriate powers of x and y, we see that:

(x+y)^6 = x^6 + 6x^5y + 15x^4y² + 20x³y³ + 15x²y^4 + 6xy^5 + y^6.

Of course you may point out that we are trying to expand (x-y)^6, not (x+y)^6. Fortunately this is easy to do, just substitute -y for y in the above expansion, and you get:

(x-y)^6 = x^6 + 6x^5(-y) + 15x^4(-y)² + 20x³(-y)³ + 15x²(-y)^4 + 6x(-y)^5 + (-y)^6.

And multiplying out the negatives:

(x-y)^6 = x^6 - 6x^5y + 15x^4y² - 20x³y³ + 15x²y^4 - 6xy^5 + y^6.

So basically, all you had to do was flip every other sign.

Answer 2

can't be solved
use binomial to expand
x^6 - 6x^5 y^1 + 15 x^4y^2 - 20 x^3y^3 + etc

Answer 3

use the binomal expansion theorem. check the source link for details.

Answer 4

have u heard of foil? first outside inside last

Answer 5

you just multiply it out

Answer 6

You don't solve an expression. Please state the problem as it was presented to you.