To factorize, we must separate the highest common factors between the products that make up the given expression. To get the highest common factor between the two products,

-24a3b3c3 = -2 * 2 *2 * 3 * a³ *b² *b * c² * c

- 84a4b2c = -2 * 2 *3 * 7 * a³ * a *b² * c

The common elements are -2, 2, 3, a³, b², c

The product of the common elements

= -12a³b²c

Hence, factorizing

-24a3b3c3 - 84a4b2c = -12a³b²c ( 2bc² + 7a)

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Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))

Marrrta [24]

Answer:

$\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Step-by-step explanation:

$\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine $\frac{1}{c} + \frac{1}{h}$

$\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine the bottom, too.

$=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}$

Apply the fraction rule

$=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}$

Cancel

$=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Therefore, $\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$