Question
Find the cube of: $a-\frac{1}{a}+b$
✓
Answer
Using $(a+b+c)^3 $
$ =a^3+b^3+c^3+3 a^2 b+3 a^2 c+3 b^2 a+3 c^2 a+6 a b c$
$\left(a-\frac{1}{a}+b\right) $
$ = a^3+\left(-\frac{1}{a}\right)^3+b^3+3 a^2\left(-\frac{1}{a}\right)+3 a^2+3\left(-\frac{1}{a}\right)^2 b+3\left(-\frac{1}{a}\right)^2 a+3 b^2a+3 b^2\left(-\frac{1}{a}\right)+6 a\left(-\frac{1}{a}\right) b$
$=a^3-\frac{1}{a^3}+b^3-3 a+3 a^2 b+\frac{3 b}{a^2}+\frac{3}{a}+3 b^2 a-\frac{3 b^2}{a}-6 b .$
$ =a^3+b^3+c^3+3 a^2 b+3 a^2 c+3 b^2 a+3 c^2 a+6 a b c$
$\left(a-\frac{1}{a}+b\right) $
$ = a^3+\left(-\frac{1}{a}\right)^3+b^3+3 a^2\left(-\frac{1}{a}\right)+3 a^2+3\left(-\frac{1}{a}\right)^2 b+3\left(-\frac{1}{a}\right)^2 a+3 b^2a+3 b^2\left(-\frac{1}{a}\right)+6 a\left(-\frac{1}{a}\right) b$
$=a^3-\frac{1}{a^3}+b^3-3 a+3 a^2 b+\frac{3 b}{a^2}+\frac{3}{a}+3 b^2 a-\frac{3 b^2}{a}-6 b .$
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