Question

(a-b)⁴-a+b pls help
thx​

Answer 1

Step-by-step explanation:

Rather than simply expanding simply use binomial theorem

Since you are looking for the patterns in binomial expansions, look at the patterns I will develop for you for (a - b)⁴. This pattern continues for any power of ( a - b)^n.

Look at Pascal's Triangle:

. . . . . 1 <=coefficients of (a + b)^0

. . . . .1 1 <=coefficients of (a + b)^1

. . . . 1 2 1 <=coefficients of (a + b)^2

. . . .1 3 3 1 <=coefficients of (a + b)^3

. . . 1 4 6 4 1 <=coefficients of (a + b)^4 <=You need these for (a - b)⁴

Notice these numbers always start and end with one, increase and then decrease with the same numbers. The center numbers can be found for any line by taking the line above it and adding the closest 2 numbers together for each spot.

. . . .1 3 . 3 .1 <=coefficients of (a + b)^3

. . . . .\ / \ / \ /

. . . .1 4 .6 .4 1 <=coefficients of (a + b)^4

When your problem has a minus in it, like (a - b)⁴, the signs in your answer will alternate, positive, then negative, then positive, then negative, etc.

If you happen to have a TI graphing calculator or its equivalent, you can get the coefficients for an expression like (a - b)⁴ by going to the Y1 = and typing in your exponent 4 (or whatever it is). Then press, MATH, PRB, and select nCr. Then type an X so your equation is now:

Y1 = 4nCrX

Look at the Table for values of this from the calculator starting when x = 0. You will see:

_x__y_

0 . . 1

1 . . 4

2 . . 6

3 . . 4

4 . . 1

The y values are the numerical coefficients of (a + b)⁴. Since their signs alternate in the equation,for (a - b)⁴ they would actually be:

_x__y_

0 . . 1

1 . .-4

2 . . 6

3 . .-4

4 . . 1

The exponent of "a" starts at 4 or your given exponent in the first term and decreases by one in each succeeding term. The exponent of "b" starts at 0 in the first term so it's not really there, and then increases by one in each succeeding term.

So,

(a - b)⁴ = a⁴- 4a³b + 6a²b² - 4ab³ + b⁴

Thank You :)