coefficent in binomial expansions: coeff. of x^2y in expansion of (x - y)^3

[helpme2457]

Pls help me

What is the co-efficient of x²y in expansion of (x - y)³

 

[Otis]

What is the co-efficient of x²y in [the] expansion of (x - y)³

Have they shown you Pascal's Triangle? That's where you get the coefficients. (If you don't know how to construct Pascal's Triangle, yet, then google it.)

Did you write out the terms in the expanded version (without coefficients), using the expansion pattern? If not, do it now.

Then, write out Pascal's Triangle far enough that you see a row containing the needed number of coefficients.

Assign the coefficients from Pascal's Triangle to your expanded polynomial, working from left-to-right, remembering that they alternate in sign.

If you would like more help, please show us what you've done so far, so that we may see where you're stuck.

 

[pka]

Pls help me
What is the co-efficient of x²y in expansion of (x - y)³

You really need to learn to cube a binomial in your head.
\(\displaystyle (a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\). I did that from memory.

You do \(\displaystyle (a+b)^3\).

 

[Steven G]

Pls help me

What is the co-efficient of x²y in expansion of (x - y)³

The 2nd term in is always + OR - the exponent in the given problem and x²y is the 2nd term. Use the sign in the middle of (x - y)³. So the answer is -3.
Of course to see that this will ALWAYS work for the 2nd term you should look at the 2nd entry of each line of Pascal's triangle.
Ans PKA is absolutely correct that you should just know how to cube a binomial in your head.