Question

Find the cube of : $x-\frac{1}{x}+y(x\ne 0)$

• A. $x^3-7x+\frac{3}{x}-\frac{1}{x^3}+3x^{2}y+\frac{3y}{x^2}-6y+3x{y}^2-\frac{3y^2}{x}+y^3$
• B. $x^3-3x+\frac{3}{x}-\frac{1}{x^3}+3x^{2}y+\frac{3y}{x^2}-6y+3x{y}^2-\frac{3y^2}{x}+y^3$
• C. $x^3-4x+\frac{3}{x}-\frac{1}{x^3}+3x^{2}y+\frac{3y}{x^2}-6y+3x{y}^2-\frac{3y^2}{x}+y^3$
• D. $x^3-9x+\frac{3}{x}-\frac{1}{x^3}+3x^{2}y+\frac{3y}{x^2}-6y+3x{y}^2-\frac{3y^2}{x}+y^3$

Answer 1

The correct option is B $x^3-3x+\frac{3}{x}-\frac{1}{x^3}+3x^{2}y+\frac{3y}{x^2}-6y+3x{y}^2-\frac{3y^2}{x}+y^3$

$(x-\frac{1}{x}+y{)}^3$

We know that

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc$

$=(x-\frac{1}{x}+y)(x^2+\frac{1}{x^2}+y^2-x(\frac{1}{-x})-xy-(\frac{1}{-x}\left)y\right)+3x\left(\frac{1}{-x}\right)y$

$=(x-\frac{1}{x}+y)(x^2+\frac{1}{x^2}+y^2+1-xy+(\frac{y}{x}\left)\right)-3y$

$=x^3-3x+\frac{3}{x}-\frac{1}{x^3}+3x^{2}y+\frac{3y}{x^2}-6y+3x{y}^2-\frac{3y^2}{x}+y^3$