Standard Identities in Algebra

What is ${\left(a+b\right)}^3$?

If ${\left(a+\frac{1}{a}+2\right)}^2=4,$ then find $a^2+\frac{1}{a^2}$, given that $\left(a^2+\frac{1}{a^2}\right)$ cannot be negative.

Without actually calculating the cubes, find the value of ${\left(-10\right)}^3+{\left(3\right)}^3+{\left(7\right)}^3$

Factories:

$125p^3+8q^3+r^3-30pqr$

Factorize:

$64x^3+27y^3+8z^3-72xyz$

Factories:

$729x^3-216y^3$

Factorizes :

$343p^3-125q^3$

Factorize:

$512a^3+216b^3$

Expand:

${\left[\frac{1}{5}x+\frac{3}{2}y\right]}^3$

Expand:

${\left[\frac{1}{4}E-\frac{1}{5}F\right]}^3$

Expand:

${\left[\frac{5}{6}A-\frac{1}{2}B\right]}^3$

Expand:

${\left[\frac{2}{5}x+\frac{1}{2}y\right]}^3$

Expand:

${\left[\frac{2}{7}x+\frac{3}{4}y\right]}^3$

Expand:

${\left[\frac{4}{5}p-\frac{7}{3}q\right]}^3$

Expand:

${\left[\frac{2}{3}x-y\right]}^3$

Expand ${\left[-3x+2y-5z\right]}^2$

Expand ${\left[-\frac{1}{3}x+\frac{1}{2}y-\frac{1}{5}z\right]}^2$

Factorize:

$\frac{1}{16}{a}^2+\frac{1}{9}{b}^2+4-\frac{1}{6}ab-\frac{4}{3}b+a$

Expand ${\left[\frac{1}{4}a-\frac{1}{3}b+2\right]}^2$

Evaluate $115\times 105$ without multiplying directly.