How do you simplify ${(x - 3)}^{3}$ $(x-3)^3$ ?

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How do you simplify ${(x - 3)}^{3}$ $(x-3)^3$ ?

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Answer 1 · 2017-05-05T08:53:52

$x^{3} - 9 x^{2} + 27 x - 27$ $x^3-9x^2+27x-27$

Explanation:

$Given (x + a) (x + b) (x + c) then expansion is$ $"Given " (x+a)(x+b)(x+c)" then expansion is"$

$x^{3} + (a + b + c) x^{2} + (a b + b c + a c) x + a b c$ $x^3+(a+b+c)x^2+(ab+bc+ac)x+abc$

$\Rightarrow {(x - 3)}^{3}$ $rArr(x-3)^3$

$= (x - 3) (x - 3) (x - 3)$ $=(x-3)(x-3)(x-3)$

$with a = b = c = - 3$ $"with " a=b=c=-3$

$\Rightarrow {(x - 3)}^{3}$ $rArr(x-3)^3$

$= x^{3} + (- 3 - 3 - 3) x^{2} + (9 + 9 + 9) x$ $=x^3+(-3-3-3)x^2+(9+9+9)x$
$\times + (- 3) (- 3) - 3)$ $color(white)(xx)+(-3)(-3)-3)$

$= x^{3} - 9 x^{2} + 27 x - 27$ $=x^3-9x^2+27x-27$

Answer 2 · 2017-05-06T09:53:05

An alternate way to work it using binomial expansion

Explanation:

An alternate way to do this is to use Binomial Expansion, which uses the general formula of:

$(a+b)^n=(C_(n,0))a^nb^0+(C_(n,1))a^(n-1)b^1+...+(C_(n,n))a^0b^n$

So here we have:

$a=x$
$b=-3$
$n=3$

$((C,a,b,"term"),(1,x^3,1,x^3),(3,x^2,-3,-9x^2),(3,x,9,27x),(1,1,-27,-27))$

and we add them up:

$x^3-9x^2+27x-27$

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