Simplify each of the following:
$(x+\frac{2}{x})^{3}+(x-\frac{2}{x})^{3}$
Given:
$(x+\frac{2}{x})^{3}+(x-\frac{2}{x})^{3}$
To do:
We have to simplify the given expression.
Solution:
We know that,
$(a+b)^3=a^3 + b^3 + 3ab(a+b)$
$(a-b)^3=a^3 - b^3 - 3ab(a-b)$
Therefore,
$(x+\frac{2}{x})^{3}+(x-\frac{2}{x})^{3}=[x^{3}+\frac{8}{x^{3}}+3 \times x^{2} \times \frac{2}{x}+3 \times x \times \frac{4}{x^{2}}]+[x^{3}-\frac{8}{x^{3}}-3 \times x^{2} \times \frac{2}{x}+3 \times x \times \frac{4}{x^{2}}]$
$=x^{3}+\frac{8}{x^{3}}+6 x+\frac{12}{x}+x^{3}-\frac{8}{x^{3}}-6 x+\frac{12}{x}$
$=2 x^{3}+\frac{24}{x}$
Hence, $(x+\frac{2}{x})^{3}+(x-\frac{2}{x})^{3}=2 x^{3}+\frac{24}{x}$.
Related Articles
- Simplify each of the following:$(\frac{x}{2}+\frac{y}{3})^{3}-(\frac{x}{2}-\frac{y}{3})^{3}$
- Simplify each of the following products:\( (x^{3}-3 x^{2}-x)(x^{2}-3 x+1) \)
- Simplify each of the following products:\( (\frac{x}{2}-\frac{2}{5})(\frac{2}{5}-\frac{x}{2})-x^{2}+2 x \)
- Solve for $x$:\( \frac{3 x-2}{3}+\frac{2 x+3}{2}=x+\frac{7}{6} \)
- Simplify each of the following expressions:\( (x+y+z)^{2}+\left(x+\frac{y}{2}+\frac{z}{3}\right)^{2}-\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\right)^{2} \)
- Solve for $x$:$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}, x≠2, 4$
- Solve the following:\( \frac{3}{4}(7 x-1)-\left(2 x-\frac{1-x}{2}\right)=x+\frac{3}{2} \)
- Which one of the following is a polynomial?(A) $\frac{x^{2}}{2}-\frac{2}{x^{2}}$(B) $\sqrt{2 x}-1$(C) $ x^{2}+\frac{3 x^{\frac{3}{2}}}{\sqrt{x}}$
- Solve for $x$:$\frac{1}{x}+\frac{2}{2x-3}=\frac{1}{x-2}, x≠0, \frac{3}{2}, 2$
- Solve the following quadratic equation by factorization: $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}, x≠2, 4$
- Solve for x:$\frac{1}{( x-1)( x-2)} +\frac{1}{( x-2)( x-3)} =\frac{2}{3} \ ,\ x\neq 1,2,3$
- Solve:\( \frac{3 x}{5}+4+x-2=\frac{\frac{3 x}{5} \times x}{2} \)
- If \( x+\frac{1}{x}=3 \), calculate \( x^{2}+\frac{1}{x^{2}}, x^{3}+\frac{1}{x^{3}} \) and \( x^{4}+\frac{1}{x^{4}} \).
- Solve the following quadratic equation by factorization: $\frac{1}{x\ -\ 3}\ +\ \frac{2}{x\ -\ 2}\ =\ \frac{8}{x};\ x\ ≠\ 0,\ 2,\ 3$
- Take away:\( \frac{6}{5} x^{2}-\frac{4}{5} x^{3}+\frac{5}{6}+\frac{3}{2} x \) from \( \frac{x^{3}}{3}-\frac{5}{2} x^{2}+\frac{3}{5} x+\frac{1}{4} \)
Kickstart Your Career
Get certified by completing the course
Get Started