Question

Proof that,
$1.{(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$
$2. {(a - b)}^{2} = ( {a}^{2} - 2ab + {b}^{2} )$
$3.(a + b)(a - b) = ( {a}^{2} - {b}^{2} )$
​

Answer 1

$solution -$

$\bf {1. {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}}$

$\bf {Proof = > we \: have,}$

$\bf {{(a + b)}^{2} = (a + b)(a + b)}$

$\bf { = a(a + b) + b(a + b)}$

$\bf { = {a}^{2} + ab + ba + {b}^{2}}$

$\bf { = {a}^{2} + 2ab + {b}^{2} [since \: ba = ab]}$

$\bf {∴ \: {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}}$

___________________________________________

$\bf {2. {(a - b)}^{2} = ( {a}^{2} - 2ab + {b}^{2} )}$

$\bf {Proof = > we \: have,}$

$\bf { {(a - b)}^{2} = (a - b)(a - b)}$

$\bf { = a(a - b) - b(a - b)}$

$\bf { = {a}^{2} - ab - ba + {b}^{2}}$

$\bf { = {a}^{2} - ab - ab + {b}^{2} [since \: ba = ab]}$

$\bf { = {a}^{2} - 2ab + {b}^{2}}$

$\bf {∴ {(a - b)}^{2} = ( {a}^{2} - 2ab + {b)}^{2}}$

___________________________________________

$\bf {3.(a + b)(a - b) = ( {a}^{2} - {b}^{2} )}$

$\bf {Proof = > we \: have,}$

$\bf {( a+ b)(a - b) = a(a - b) + b(a - b)}$

$\bf { = {a}^{2} - ab + ba - {b}^{2}}$

$\bf { = {a}^{2} - ab + ab - {b}^{2} [since \: ba = ab)]}$

$\bf { = {a}^{2} - {b}^{2}}$

$\bf { ∴(a + b)(a - b) = ( {a}^{2} - {b}^{2} )}$

____ ____ ____ ____ ___ ____ ____ ____....