Question

Find the cube of:

$\left(vi\right)(a-\frac{1}{a}+b)$

Answer 1

Given Expression is $(a-\frac{1}{a}+b)$
Now, find the cube of $(a-\frac{1}{a}+b)$ using the formula $(a+b+c{)}^3=a^3+b^3+c^3+3a^{2}b+3a^{2}c+3b^{2}a+3b^{2}c+3c^{2}a+3c^{2}b+6abc$

$\Rightarrow {(a-\frac{1}{a}+b)}^3$

$=a^3+{(-\frac{1}{a})}^3+b^3+3a^2(-\frac{1}{a})+3a^{2}b+3{(-\frac{1}{a})}^{2}a+3{(-\frac{1}{a})}^{2}b+3b^{2}a+3b^2(-\frac{1}{a})+6a(-\frac{1}{a})b$

$=a^3-\frac{1}{a^3}+b^3-3a+3a^{2}b+\frac{3}{a}+\frac{3b}{a^2}+3b^{2}a-\frac{3b^2}{a}-6b$

Hence, the cube of $(a-\frac{1}{a}+b)$ is $a^3-\frac{1}{a^3}+b^3-3a+3a^{2}b+\frac{3}{a}+\frac{3b}{a^2}+3b^{2}a-\frac{3b^2}{a}-6b$