Question

4(x+3y)³ + 12(x+3y)²​

Answer 1

Answer:

$4\left(x+3y\right)^3+12\left(x+3y\right)^2\\= 4\left(x^3+9x^2y+27xy^2+27y^3\right)+12\left(x+3y\right)^2\\= 4\left(x^3+9x^2y+27xy^2+27y^3\right)+12\left(x^2+6xy+9y^2\right)\\= 4x^3+36x^2y+108xy^2+108y^3+12\left(x^2+6xy+9y^2\right)\\= 4x^3+36x^2y+108xy^2+108y^3+12x^2+72xy+108y^2$

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Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

Before solving question look at this identities :

$\sf {\boxed{ {(x +y)}^{2} = {x}^{2} + {y}^{2} +2xy}}$

$\sf\boxed{{(x + y)}^{3} = {x}^{3} + {y}^{3} + 3 {x}^{2} y + 3x {y}^{2}}$

$\sf = > 4[( {x)}^{3} + {(3y)}^{3} + 3 {x}^{2} y + 3x {(3y)}^{2} ]+ 12[{x}^{2} + {(3y)}^{2} + 2(x)(3y)]$

$\sf = > 4[{x}^{3} + 2 7{y}^{3} + 9 {x}^{2} y + 27x {y}^{2} ]+ 12( {x}^{2} + 9 {y}^{2} + 6xy)$

$\sf = > 4 {x}^{3} + 108 {y}^{3} + 36 {x}^{2} y + 108x {y}^{2} + 12 {x}^{2} + 108 {y}^{2} + 72xy$

$\sf = > 4 {x}^{3} + 108 {y}^{3} + 108x {y}^{2} + 108 {y}^{2} + 36 {x}^{2} y + 12 {x}^{2} + 72xy$

Extra information=>

Other Identities:

$\sf \bold{\boxed{ {(x -y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\sf \boxed{{x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}$

$\sf \bold{\boxed{(x + y)(x + z) = {x}^{2} + (y+ z)x + yz}}$

$\bold{\boxed{ {(x -y)}^{3} = {x}^{3} + {y}^{3} -3xy[x-y]}}$

$\sf \bold{\boxed{(x + a)(x - b) = {x}^{2} + (a - b)x - ab}}$

$\sf \bold{\boxed{(x - a)(x - b) = {x}^{2} - (a + b)x + ab}}$