Question

Prove that: (a + b)-1 (a-1 + b-1) = (ab)-1.​

Answer 1

Answer:

answer

Step-by-step explanation:

here. is your answers

Answer 2

Answer:

Given,

${(a + b)}^{ - 1} ( {a}^{ - 1} + {b}^{ - 1})$

This is a problem of Algebra,

${(a + b)}^{ - 1} ( {a}^{ - 1} + {b}^{ - 1}) = \frac{1}{a + b} ( \frac{1}{a} + \frac{1}{b} )$

$= \frac{1}{(a + b)} ( \frac{b + a}{ab} )$

$= \frac{1}{ab}$

$= {(ab)}^{ - 1}$

Therefore,

${(a + b)}^{ - 1} ( {a}^{ - 1} + {b}^{ - 1}) = {(ab)}^{ - 1}$

[proved]

Some extra formulas of Algebra,

1.(a + b)² = a² + 2ab + b²

2.(a − b)² = a² − 2ab − b²

3.(a + b)³ = a³ + 3a²b + 3ab² + b³

4.(a - b)³ = a³ - 3a²b + 3ab² - b³

5.(a³ + b³) = (a + b)³ − 3ab(a + b)

6.(a³ - b³)= (a -b)³ + 3ab(a - b)

$7.{a}^{2} - {b}^{2} = (a + b)(a - b) \\ 8.{a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab \\ 9.{a}^{2} + {b}^{2} = {(a - b)}^{2} + 2ab \\ 10.{a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} ) \\ 11.{a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )$