What's $1/7-(1/7)^2+(1/7)^3$ ?

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Answer 1 · 2017-03-14T01:28:35

$43/343$

We can combine fractions like this:

$1/7-(1/7)^2+(1/7)^3$

$1/7-1^2/7^2+1^3/7^3=1/7-1/7^2+1/7^3$

Now let's find a common denominator:

$1/7(1)-1/7^2(1)+1/7^3$

$1/7(7^2/7^2)-1/7^2(7/7)+1/7^3$

$7^2/7^3-7/7^3+1/7^3$

$(7^2-7+1)/7^3=(49-7+1)/343=43/343$

What's 1/7-(1/7)^2+(1/7)^31/7-(1/7)^2+(1/7)^3?