Question

(3a+b)^3 + (3a-b)^3​

Answer 1

$\huge\tt{\bold{\underline{\underline{Question᎓}}}}$

(3a+b)^3 + (3a-b)^3

$\huge\tt{\bold{\underline{\underline{Answer᎓}}}}$

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Here ,this identity is used:-

$\bold{\red{ = > {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)}}$

$\bold{\pink{= > {(a - b)}^{3} = {a}^{3} + {b}^{3} - 3ab(a - b)}}$

Now come to the Question:-

$= > {(3a + b)}^{3} = {(3a)}^{3} + {b}^{3} + 3(3a)(b)(3a + b)$

$= 27 {a}^{3} + {b}^{3} + 9ab(3a + b)$

$= 27 {a}^{3} + {b}^{3} + 27 {a}^{2} b + 9a {b}^{2}$

$= > {(3a - b)}^{3} = {(3a)}^{3} + {b}^{3} - 3(3a)(b)(3a - b)$

$= 27 {a}^{3} + {b}^{3} - 9ab(3a - b)$

$= 27 {a}^{3} + {b}^{3} - 27 {a}^{2} b + 9a {b}^{2}$

$\bold{Now \:,adding \:both\: terms }$

$= > {(3a + b)}^{3} + {(3a - b)}^{3}$

$= 27 {a}^{3} + {b}^{3} + 27 {a}^{2} b + 9a {b}^{2} + 27 {a}^{3} + {b}^{3} - 27 {a}^{2} b + 9a {b}^{2}$

$= > 27 {a}^{3} + {b}^{3} + 9a {b}^{2} + 27 {a}^{3} + {b}^{3} + 9a {b}^{2}$

$= > 54 {a}^{3} + 2 {b}^{3} +18a {b}^{2}$

=>Taking out 2 common from whole equation:-

$\bold{= > 2(27 {a}^{3} + {b}^{3} + 9a {b}^{2} )}$

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Answer 2

above answer is correct mark them brainliest :-