Question

Question #78619

Answer 1

`(x-7)^3= x^3 - 21x^2 + 147x -343`

Explanation:

For low powers you can make use of Pascal's Triangle but I'll do it more generally here with the binomial theorem:

`(x+y)^n = sum_(k=0)^n ((n),(k))x^(n-k) y^k`

where `((n),(k)) = (n!)/((n-k)!k!`

Hence, we have

`(x-7)^3 = sum_(k=0)^3 ((3),(k)) x^(3-k) y^k`

`=((3),(0))x^3 + ((3),(1))x^2 (-7) + ((3),(2))x (-7)^2 + ((3),(3)) (-7)^3`

we evaluate this, noting that `0! = 1`

`= x^3 - 21x^2 + 147x -343`

Answer 2

`x^3-21x^2+147x-343`

Explanation:

`"to expand "(x+a)^3" in general"`

`x^3+(a+a+a)x^2+(a^2+a^2+a^2)x+a^3`

`"here "a=-7`

`rArr(x-7)^3`

`=x^3+(-7-7-7)x^2+(49+49+49)x+(-7)^3`

`=x^3-21x^2+147x-343`