Chapter 7.1, Problem 55PPS

To determine

To simplify:An expression ${\left(-3r^3{w}^4\right)}^3{\left(2rw\right)}^2{\left(-3r^2\right)}^3{\left(4r{w}^2\right)}^3{\left(2r^2{w}^3\right)}^4$ .

Answer to Problem 55PPS

The required answer is $2985984r^{28}{w}^{32}$ .

Explanation of Solution

Given information:

The given expression is ${\left(-3r^3{w}^4\right)}^3{\left(2rw\right)}^2{\left(-3r^2\right)}^3{\left(4r{w}^2\right)}^3{\left(2r^2{w}^3\right)}^4$ .

Calculation:

The given expression:

$={\left(-3r^3{w}^4\right)}^3{\left(2rw\right)}^2{\left(-3r^2\right)}^3{\left(4r{w}^2\right)}^3{\left(2r^2{w}^3\right)}^4$

$={\left(-3\right)}^3{\left(r^3\right)}^3{\left(w^4\right)}^3{2}^2{r}^2{w}^2{\left(-3\right)}^3{\left(r^2\right)}^3{4}^3{r}^3{\left(w^2\right)}^3{2}^4{\left(r^2\right)}^4{\left(w^3\right)}^4$

(Power of a product)

$=-27\cdot r^9\cdot w^{12}\cdot 4\cdot r^2\cdot w^2\left(-27\right)r^6\cdot 64\cdot r^3\cdot w^6\cdot 16\cdot r^8\cdot w^{12}$

(Power of a power and ${\left(-3\right)}^3=-27;2^2=4;4^3=64;2^4=16$ )

$=\left[\left(-27\cdot 4\cdot 64\cdot 16\right)\left(-27\right)\right]\left(r^9\cdot r^2\cdot r^6\cdot r^3\cdot r^8\right)\left(w^{12}\cdot w^2\cdot w^6\cdot w^{12}\right)$

(Commutative)

$=2985984\left(r^{9+2+6+3+8}\right)\left(w^{12+2+6+12}\right)$

(Product of powers)

$=2985984r^{28}{w}^{32}$

(Simplify)

So, the required answer is $2985984r^{28}{w}^{32}$ .