Question

can anyone help me?
http://i.imgur.com/DivUi7D.jpg

Answer 1

try factoring bottom, using : \( a^3-b^3 = (a-b)(a^2+ab+b^2)\)

Answer 2

wait, the denominator? x-1? u can do that? D:

Answer 3

\(\large x-1 = (\sqrt[3]{x})^3 - 1^3 \)

Answer 4

ohhh

Answer 5

\[(a ^{3}-b ^{3})=(a-b)(a ^{2}+ab+ b ^{2})\]a = cube root x and b = 1
\[(\sqrt[3]{x}-1)(\sqrt[3]{x ^{2}}+\sqrt[3]{x}+1)\]

Answer 6

oh yay :D i got -1/3

Answer 7

Bingo :3

Answer 8

thanks guys :D!

Answer 9

Yep yep