Question

calcule: $(x-2) ^{3}$

Answer 1

Observe que:

$(a-b)^3=a^3-3a^2b+3ab^2-b^3$

Logo:

$(x-2)^3=x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3$

$\boxed{(x-2)^3=x^3-6x^2+12x-8}$

Pelo Binômio de Newton:

$(a-b)^{n}=\dbinom{n}{0}a^nb^0-\dbinom{n}{1}a^{n-1}b^1+\dots+\dbinom{n}{n-1}a^1b^{n-1}-\dbinom{n}{n}a^0b^n$

Assim, $(x-2)^3=\dbinom{3}{0}x^3-\dbinom{3}{1}x^2\cdot2+\dbinom{3}{2}x\cdot2^2-\dbinom{3}{3}2^3$.

Como $\dbinom{3}{0}=1$, $\dbinom{3}{1}=3$, $\dbinom{3}{2}=3$ e $\dbinom{3}{3}=1$, segue que:

$(x-2)^3=x^3-3x^2\cdot2+3x\cdot2^2-2^3$

$\boxed{(x-2)^3=x^3-6x^2+12x-8}$