Question

Please provide answer for the attached question with proper steps.
Please provide with proper steps.

Answer 1

Given :-

$ab(a - b)(a + b)( {a}^{2} + {b}^{2} ) = {a}^{m} {b}^{n} - {a}^{n} {b}^{m}$

To find :-

m - n

(a-b)(a+b) = a² - b² [Identity]

Therefore,

$ab( {a}^{2} - {b}^{2})( {a}^{2} + {b}^{2}) = {a}^{m} {b}^{n} - {a}^{n} {b}^{m}$

The identity can be applied for (a²-b²)(a²+b²) :-

a⁴ - b⁴ = (a²)² - (b²)²

=≥ (a²-b²)(a²+b²)

$ab( {a}^{4} - {b}^{4}) = {a}^{m} {b}^{n} - {a}^{n} {b}^{m}$

$[(a \times b \times {a}^{4}) - (a \times b \times {b}^{4})] = {a}^{m} { b}^{n} - {a}^{n} {b}^{m}$

By multiplying ab with a⁴-b⁴

${a}^{5}b - a {b}^{5} = {a}^{m} {b}^{n} - {a}^{n} {b}^{m}$

Therefore we can conclude that

m = 5

n = 1

Hence,

m - n = 5 - 1 =≥ 4

Option (3) 4 is correct

Important points to remember :-

${a}^{m} \times {a}^{n} = {a}^{m + n}$

${a}^{m} \times {b}^{m} = {ab}^{m}$

$\dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$\dfrac{ {a}^{m} }{ {b}^{m} } = (\dfrac{a}{b})^{m}$

${a}^{1} = a$

${a}^{0} = 1$

$\sqrt{a} = {a}^{ \frac{1}{2} }$