Question #78619

2 Answers
Aug 24, 2017

(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

Aug 24, 2017

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