Question

(x+y) ^3-(x-y)^3 can be factorizatized as?

Answer 1

$We \:know \:the \: Algebraic \: Identity$

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

$Here, a = x + y , \: and \: b = x - y$

$Now , (x+y)^{3} - (x-y)^{3} \\= [(x+y)-(x-y)]^{3} + 3(x+y)(x-y)[(x+y)-(x-y)] \\= ( x+y-x+y)^{3} + 3(x^{2} -y^{2} )( x+y-x+y) \\= (2y)^{3} +3(x^{2}-y^{2})\times 2y \\= 2y[ (2y)^{2} +3(x^{2} -y^{2} )] \\= 2y[ 4y^{2} + 3x^{2}-3y^{2} ]\\= 2y( 3x^{2} + y^{2})$

Therefore.,

$\red{ Factors \:of \: (x+y)^{3} - (x-y)^{3}} \\\green { =2y( 3x^{2} + y^{2} )}$

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