Use Pascal's triangle to expand (a+b)^6 and hence find the binomial expansion of: (x-3)^6

I can not do the symmetry with keyboard

remember rows and columns start wit 0, so we need 7 rows
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 1
1 6 15 20 15 6 1 ah ha
so
a^6 + 6 a^5 b + 15 a^4 b^2 + 20 a^3 b^3 + 15 a^2 b^4 + 6 a b^5 + b^6

I hope it was right

To expand (a + b)^6 using Pascal's triangle, we can use the coefficients from the 6th row of Pascal's triangle. The 6th row of Pascal's triangle is:

1 5 10 10 5 1

The coefficients represent the coefficients of each term when expanding (a + b)^6. The exponents of a decrease by one for each term, starting from 6, while the exponents of b increase by one for each term, starting from 0.

So, the binomial expansion of (a + b)^6 is:

1a^6 + 5a^5b + 10a^4b^2 + 10a^3b^3 + 5a^2b^4 + 1ab^5 + 1b^6

Now let's use the binomial expansion to find (x - 3)^6.

By using the same coefficients from the 6th row of Pascal's triangle, we can substitute x for a and -3 for b in the binomial expansion of (a + b)^6.

So, the binomial expansion of (x - 3)^6 is:

1x^6 + 5x^5(-3) + 10x^4(-3)^2 + 10x^3(-3)^3 + 5x^2(-3)^4 + 1x(-3)^5 + 1(-3)^6

Simplifying each term, we have:

x^6 - 15x^5 + 90x^4 - 270x^3 + 405x^2 - 243x + 729

Therefore, the binomial expansion of (x - 3)^6 is:

x^6 - 15x^5 + 90x^4 - 270x^3 + 405x^2 - 243x + 729

To expand (a+b)^6 using Pascal's triangle, we need to know the coefficients of the binomial expansion. Pascal's triangle provides a convenient way to determine these coefficients.

Pascal's triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. The first row is simply a 1, and each subsequent row starts and ends with a 1. Here is a portion of Pascal's triangle:

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

To expand (a+b)^6, we need to look at the sixth row of Pascal's triangle. The coefficients of the terms in the expansion can be found by reading the numbers from left to right in the sixth row:

1 5 10 10 5 1

These coefficients correspond to the powers of a and b in the binomial expansion. The first term has a coefficient of 1, which corresponds to a^6. The second term has a coefficient of 5, which corresponds to a^5b. The third term has a coefficient of 10, which corresponds to a^4b^2. And so on.

Now let's use this expansion to find the binomial expansion of (x-3)^6:

(x-3)^6 = 1(x^6) + 5(x^5)(-3) + 10(x^4)(-3)^2 + 10(x^3)(-3)^3 + 5(x^2)(-3)^4 + 1(x)(-3)^5 + 1(-3)^6

Simplifying each term, we get:

(x-3)^6 = x^6 - 15x^5 + 90x^4 - 270x^3 + 405x^2 - 324x + 729

So the binomial expansion of (x-3)^6 is:

(x-3)^6 = x^6 - 15x^5 + 90x^4 - 270x^3 + 405x^2 - 324x + 729