Question

How do you expand `(3a – b)^6`?

Answer 1

`(3a-b)^6 = 729a^6 - 1458a^5b + 1215a^4b^2 + -540a^3b^3 + 135a^2b^4 -18 ab^5 + b^6`

Explanation:

Recall the Binomial Theorem

`(a+b)^n = sum_(r=0)^n \ ( (n), (r) ) \ a^(n-r) \ b^r`

Where:

`( (n), (r) ) =""^nC_r = (n!)/((n-r)!r!`

is the combinatorial, which are also the numbers from the `nth` row of Pascal's Triangle

Thus we have:

`(3a-b)^6 = sum_(r=0)^6 \ ( (6), (r) ) \ (3a)^(5-r) \ (-b)^r`

`" " = ( (6), (0) ) (3a)^(6)(-b)^(0) + ( (6), (1) ) (3a)^(5)(-b)^(1) + ( (6), (2) ) (3a)^(4)(-b)^(2) + ( (6), (3) ) (3a)^(3)(-b)^(3) + ( (6), (4) ) (3a)^(2)(-b)^(4) + ( (6), (5) ) (3a)^(1)(-b)^(5) + ( (6), (6) ) (3a)^(0)(-b)^(6)`

`" " = (1) (729a^6) + (6) (243a^5)(-b) + (15) (81a^4)(b^2) + (20) (27a^3)(-b^3) + (15) (9a^2)(b^4) + (6) (3a)(-b^5) + (b^6)`

`" " = 729a^6 - 1458a^5b + 1215a^4b^2 + -540a^3b^3 + 135a^2b^4 -18 ab^5 + b^6`