Question

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Find the product
(i) (a+7)×(a²+3a+)
(ii)(a + b + c)(a-b-c)​

Answer 1

$\huge {\overbrace {\underbrace {\underline {\blue {answer}}}}}$

$(i)(a + 7)( {a}^{2} + 3a)$

$= (a + 7)( {a}^{2} + 3a)$

$= (a)( {a}^{2} ) + (a)(3a) +( 7 )({a}^{2} ) +( 7)(3a)$

$= {a}^{3} + 3 {a}^{2} + 7 {a}^{2} + 21a$

$= {a}^{3} + 10 {a}^{2} + 21a$

$(ii))(a + b + c)(a-b-c)$

$= (a)(a) + (a)( - b) + (a)( - c) + (b)(a) + (b)( - b) + (b)( - c) + (c)(a) + (c)( - a) + (c)$

$= {a}^{2} - ab - ac + ab - {b}^{2} - bc + ac - bc$

$= {a}^{2} - {b}^{2} - 2bc - {c}^{2}$

Answer 2

Answer:

i)(a+7)(a2+3a)

= (a + 7)( {a}^{2} + 3a)=(a+7)(a2+3a)

= (a)( {a}^{2} ) + (a)(3a) +( 7 )({a}^{2} ) +( 7)(3a)=(a)(a2)+(a)(3a)+(7)(a2)+(7)(3a)

= {a}^{3} + 3 {a}^{2} + 7 {a}^{2} + 21a=a3+3a2+7a2+21a

= {a}^{3} + 10 {a}^{2} + 21a=a3+10a2+21a

(ii))(a + b + c)(a-b-c)(ii))(a+b+c)(a−b−c)

= (a)(a) + (a)( - b) + (a)( - c) + (b)(a) + (b)( - b) + (b)( - c) + (c)(a) + (c)( - a) + (c)=(a)(a)+(a)(−b)+(a)(−c)+(b)(a)+(b)(−b)+(b)(−c)+(c)(a)+(c)(−a)+(c)

= {a}^{2} - ab - ac + ab - {b}^{2} - bc + ac - bc=a2−ab−ac+ab−b2−bc+ac−bc

= {a}^{2} - {b}^{2} - 2bc - {c}^{2}=a2−b2−2bc−c2