Could you show some work of what you've done so we know how we can help you?
Ok well, we can do it your way first.
Just continue foiling you're almost there.
(x^2 + xy + xy+ y ^2) (x^2 + xy + xy+ y ^2)
You get
(x^2 + 2xy + y^2) (x^2 + 2xy + y^2)
Just finish it up.
(x^2)(x^2)+(x^2)(2xy)+(x^2)(y^2)..........
It's long, I know but if you forget pascal's triangle during a test, this is your best hope, write neatly and legibly.
As for Pascal's triangle...
You know that it's
1
11
121
1331
And so on...
If you think about it, these are the powers of 11.
11^0 = 1
11^1 = 11
11^2 = 121
.....
That's how I remember which level I take the numbers from.
Since you're expanding to 4.
You take it from the 4th level which is 1331
I recommend you write it out this way
1 * x * y
3 * x * y
3 * x * y
1 * x * y
Now you know from (x+y)^2, you get x^2 + 2xy + y^2
You would've gotten it this way.
1 * x^2 * y^0 = 1(x^2)(y^0) = x^2
2 * x^1 * y^1 = 2(x^1)(y^1) = 2xy
1 * x^0 * y^2 = 1(x^0)(y^2) = y^2
[Then you add them all up.
x^2 + 2 xy + y^2, that's the answer from (x+y)^2]
Realize how x starts from 2 and ends at 0 whereas y starts from 0 and ends at 2.
So does the same thing for this.
1 * x * y
3 * x * y
3 * x * y
1 * x * y
X starts from 4 and y starts from 0.
I hope this helps.