Use Pascal's Triangle to expand the binomial.
(d – 3)6
To expand the binomial (d – 3)^6 using Pascal's Triangle, we can use the following pattern:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
The numbers in each row of Pascal's Triangle represent the coefficients of the expanded binomial. The powers of "d" decrease from left to right, while the powers of "-3" increase. To expand (d – 3)^6, we use the coefficients from the 6th row of Pascal's Triangle, which are:
1, 6, 15, 20, 15, 6, 1
We substitute these coefficients into the equation:
(d – 3)^6 = 1(d^6) + 6(d^5)(-3) + 15(d^4)(-3)^2 + 20(d^3)(-3)^3 + 15(d^2)(-3)^4 + 6(d)(-3)^5 + 1(-3)^6
Expanding the terms and simplifying, we get:
= d^6 - 18d^5 + 135d^4 - 540d^3 + 1215d^2 - 1458d + 729
Therefore, (d – 3)^6 expands to d^6 - 18d^5 + 135d^4 - 540d^3 + 1215d^2 - 1458d + 729.
To expand the binomial (d - 3)^6 using Pascal's Triangle, we can use the binomial theorem. The binomial theorem states that the expansion of (a + b)^n can be found by summing the coefficients of the terms in the expansion, where each coefficient is given by the corresponding entry in Pascal's Triangle.
The expansion of (d - 3)^6 can be written as:
1(d^6) + 6(d^5)(-3) + 15(d^4)(-3)^2 + 20(d^3)(-3)^3 + 15(d^2)(-3)^4 + 6(d)(-3)^5 + 1(-3)^6
Using Pascal's Triangle, the coefficients are 1, 6, 15, 20, 15, 6, and 1, respectively.
Simplifying the terms further:
d^6 - 18d^5 + 135d^4 - 540d^3 + 1215d^2 - 1458d + 729
To expand the binomial (d – 3)^6 using Pascal's Triangle, we can use the binomial theorem. The binomial theorem states that for any positive integers n and m:
(a + b)^n = C(n, 0)a^n + C(n, 1)a^(n-1)b + C(n, 2)a^(n-2)b^2 + ... + C(n, m)a^(n-m)b^m + ... + C(n, n)b^n
Where C(n, m) represents the binomial coefficient, which can be found using Pascal's Triangle.
In this case, our binomial is (d – 3), and we want to expand it to the power of 6. Our term will have 7 terms in total, starting with d to the power of 6 and ending with -3 to the power of 6. Therefore, we will use the binomial coefficients from the 7th row of Pascal's Triangle.
The 7th row of Pascal's Triangle is: 1 6 15 20 15 6 1
Using these coefficients, we can write the expansion of (d – 3)^6 as:
(d – 3)^6 = C(6, 0)d^6 + C(6, 1)d^5(-3) + C(6, 2)d^4(-3)^2 + C(6, 3)d^3(-3)^3 + C(6, 4)d^2(-3)^4 + C(6, 5)d(-3)^5 + C(6, 6)(-3)^6
Now we substitute the values from the row of Pascal's Triangle:
(d – 3)^6 = 1(d^6) + 6(d^5)(-3) + 15(d^4)(-3)^2 + 20(d^3)(-3)^3 + 15(d^2)(-3)^4 + 6(d)(-3)^5 + 1(-3)^6
Simplifying each term, the expanded form becomes:
(d – 3)^6 = d^6 - 18d^5 + 135d^4 - 540d^3 + 1215d^2 - 1458d + 729