Chapter 1.3, Problem 135E

  $\left(a\right)$

To determine

To calculate: The factors of the expression $A^4-B^4\text{ and }{A}^6-B^6$ .

Answer to Problem 135E

The factors of the expression $A^4-B^4\text{=}\left(A-B\right)\left(A+B\right)\left(A^2+B^2\right)$

  $\text{and }{A}^6-B^6=\left(A-B\right)\left(A^2+B^2+AB\right)\left(A+B\right)\left(A^2+B^2-AB\right)$ .

Explanation of Solution

Given information:

The expression $A^4-B^4\text{ and }{A}^6-B^6$ .

Formula used:

The difference of square of two numbers a and b is $a^2-b^2=\left(a+b\right)\left(a-b\right)$ .

Calculation:

Consider the expression $A^4-B^4\text{ and }{A}^6-B^6$ .

Recall that the difference of square of two numbers a and b is $a^2-b^2=\left(a+b\right)\left(a-b\right)$ .

Apply it,

  $\begin{array}{c}{A}^4-B^4={\left(A^2\right)}^2-{\left(B^2\right)}^2\\ =\left(A^2-B^2\right)\left(A^2+B^2\right)\end{array}$

Again apply the difference of square of two numbers,

  $\begin{array}{c}{A}^4-B^4=\left(A^2-B^2\right)\left(A^2+B^2\right)\\ =\left(A-B\right)\left(A+B\right)\left(A^2+B^2\right)\end{array}$

And,

  $\begin{array}{c}{A}^6-B^6={\left(A^3\right)}^2-{\left(B^3\right)}^2\\ =\left(A^3-B^3\right)\left(A^3+B^3\right)\end{array}$

Again apply the difference and sum of cube of two numbers,

  $\begin{array}{c}{A}^6-B^6={\left(A^3\right)}^2-{\left(B^3\right)}^2\\ =\left(A^3-B^3\right)\left(A^3+B^3\right)\\ =\left(A-B\right)\left(A^2+B^2+AB\right)\left(A+B\right)\left(A^2+B^2-AB\right)\end{array}$

Thus, the factors of the expression $A^4-B^4\text{=}\left(A-B\right)\left(A+B\right)\left(A^2+B^2\right)$

  $\text{and }{A}^6-B^6=\left(A-B\right)\left(A^2+B^2+AB\right)\left(A+B\right)\left(A^2+B^2-AB\right)$ .

  $\left(b\right)$

To determine

To verify: The values of the expression $18335={12}^4-7^4\text{ and }2868335={12}^6-7^6$ .

Explanation of Solution

Given information:

The expression $18335={12}^4-7^4\text{ and }2868335={12}^6-7^6$ .

Formula used:

The difference of fourth power of two numbers A and B is $A^4-B^4\text{=}\left(A-B\right)\left(A+B\right)\left(A^2+B^2\right)$ and difference of sixth power $A^6-B^6=\left(A-B\right)\left(A^2+B^2+AB\right)\left(A+B\right)\left(A^2+B^2-AB\right)$ .

Calculation:

Consider the expression $18335={12}^4-7^4\text{ and }2868335={12}^6-7^6$ .

Recall that the difference of fourth power of two numbers A and B is $A^4-B^4\text{=}\left(A-B\right)\left(A+B\right)\left(A^2+B^2\right)$ and difference of sixth power $A^6-B^6=\left(A-B\right)\left(A^2+B^2+AB\right)\left(A+B\right)\left(A^2+B^2-AB\right)$ .

Apply it,

  $\begin{array}{c}{12}^4-7^4=\left(12-7\right)\left(12+7\right)\left({12}^2+7^2\right)\\ =5\left(19\right)\left(144+49\right)\\ =5\left(19\right)\left(193\right)\\ =18335\end{array}$

And,

  $\begin{array}{c}{12}^6-7^6=\left(12-7\right)\left(12+7\right)\left({12}^2+7^2+12\left(7\right)\right)\left({12}^2+7^2-12\left(7\right)\right)\\ =5\left(19\right)\left(144+49+84\right)\left(144+49-84\right)\\ =5\left(19\right)\left(277\right)\left(109\right)\\ =2868335\end{array}$

Hence, it is verified that values of the expression $18335={12}^4-7^4\text{ and }2868335={12}^6-7^6$ .

  $\left(c\right)$

To determine

To show: The factors $18335\text{ and }2868335$ are prime numbers.

Explanation of Solution

Given information:

The expression $18335={12}^4-7^4\text{ and }2868335={12}^6-7^6$ .

Formula used:

The difference of fourth power of two numbers A and B is $A^4-B^4\text{=}\left(A-B\right)\left(A+B\right)\left(A^2+B^2\right)$ and difference of sixth power $A^6-B^6=\left(A-B\right)\left(A^2+B^2+AB\right)\left(A+B\right)\left(A^2+B^2-AB\right)$ .

Calculation:

Consider the expression $18335={12}^4-7^4\text{ and }2868335={12}^6-7^6$ .

Recall that the difference of fourth power of two numbers A and B is $A^4-B^4\text{=}\left(A-B\right)\left(A+B\right)\left(A^2+B^2\right)$ and difference of sixth power $A^6-B^6=\left(A-B\right)\left(A^2+B^2+AB\right)\left(A+B\right)\left(A^2+B^2-AB\right)$ .

Apply it,

  $\begin{array}{c}18335={12}^4-7^4\\ =\left(12-7\right)\left(12+7\right)\left({12}^2+7^2\right)\\ =5\left(19\right)\left(144+49\right)\\ =5\left(19\right)\left(193\right)\end{array}$

Therefore, factors of $18335$ are 5, 19 and 193 which are all prime numbers.

And,

  $\begin{array}{c}2868335={12}^6-7^6\\ =\left(12-7\right)\left(12+7\right)\left({12}^2+7^2+12\left(7\right)\right)\left({12}^2+7^2-12\left(7\right)\right)\\ =5\left(19\right)\left(144+49+84\right)\left(144+49-84\right)\\ =5\left(19\right)\left(277\right)\left(109\right)\end{array}$

Therefore, factors of $2868335$ are 5, 19, 277 and 109 which are all prime numbers.

Hence, it is shown that factors $18335\text{ and }2868335$ are prime numbers.